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Dynamics Problems

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Warm Up The system below is in equilibrium. If the scale is calibrated in N, what does it read? 5kg Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N

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Warm Up Determine the Normal force in each of the following: m m F F F F F W W F m m

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Warm Up Write F=ma for each scenario: m m F F a F W W F m m

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Example A pumpkin of unknown mass is suspended by a cord attached to the ceiling and pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an angle of to horizontal, the pumpkin will remain in equilibrium when the cord makes an angle of with the vertical. (A) What is the tension in the cord when the pumpkin is in equilibrium ? (B) What is the mass of the pumpkin ? Diagram Free body Diagram

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(A) What is the tension in the cord when the pumpkin is in equilibrium ? Since the pumpkin is in static equilibrium, we need only look at the horizontal components

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(B) What is the mass of the pumpkin ? Since the pumpkin is in static equilibrium, we need only look at the vertical components

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Example The tension in the horizontal rope is 30N A) Determine the weight of the object 30N Diagram Free body Diagram 50 0

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A) Determine the weight of the object 50 0 The weight is in static equilibrium, so the appropriate net component forces must be zero. Horizontal Component Vertical Component The weight of the mass is 36N

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Weight on a Wire A rope extends between two poles. A 80N weight hangs from it as per the diagram. A rope extends between two poles. A 80N weight hangs from it as per the diagram. A) Determine the tension in both parts of the rope. A) Determine the tension in both parts of the rope N T1T1 T2T2 Diagram Free body Diagram

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A) Determine the tension in both parts of the rope. The weight is in static equilibrium, so the appropriate net component forces must be zero. Horizontal Component Vertical Component

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Example The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. Free body Diagram Block Ring 35N 7.0N

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The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. The weight is just in static equilibrium, so the appropriate net component forces must be zero. Block Ring From Block: From ring: Combining:

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Atwood’s Machine Example: a) What are the tensions in the string T 1 and T 2 ? b) Find the accelerations, a 1 and a 2, of the masses. Masses m 1 = 10 kg and m 2 = 20kg are attached to an ideal massless string and hung as shown around an ideal massless pulley. Fixed Pulley m1m1 m2m2 a1a1 a2a2 T1T1 T2T2

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Draw free body diagrams for each object Draw free body diagrams for each object Applying Newton’s Second Law: Applying Newton’s Second Law: T 1 - m 1 g = m 1 a 1 (a) T 1 - m 1 g = m 1 a 1 (a) T 2 - m 2 g = -m 2 a 2 T 2 - m 2 g = -m 2 a 2 => -T 2 + m 2 g = +m 2 a 2 (b) But T 1 = T 2 = T But T 1 = T 2 = T since pulley is ideal since pulley is ideal and a 1 = -a 2 =a and a 1 = -a 2 =a since the masses are since the masses are connected by the string connected by the string m2gm2g m1gm1g Free Body Diagrams T1T1 T2T2 a1a1 a2a2

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-m 1 g + T = m 1 a(a) -m 1 g + T = m 1 a(a) -T + m 2 g = m 2 a(b) -T + m 2 g = m 2 a(b) Two equations and two unknowns we can solve for both unknowns (T and a). we can solve for both unknowns (T and a). Add (b) + (a): Add (b) + (a): g(m 2 – m 1 ) = a(m 1 + m 2 ) g(m 2 – m 1 ) = a(m 1 + m 2 ) m2gm2g m1gm1g Free Body Diagrams T1T1 T2T2 a1a1 a2a2 Solve for Acceleration

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-m 1 g + T = m 1 a(a) -m 1 g + T = m 1 a(a) -T + m 2 g = m 2 a(b) -T + m 2 g = m 2 a(b) Plug a into (b) and Solve for T m2gm2g m1gm1g Free Body Diagrams T1T1 T2T2 a1a1 a2a2 Solve for T

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m1m1 m2m2 a a T T So we find: So we find: Atwood Machine Review

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Example A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

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Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? Free body Diagram +y +x a

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Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? Vertical Forces Horizontal Forces Solving +y +x a

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Example Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

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Pulling a Box (Part 2) Free body Diagram +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

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Pulling a Box (Part 2) +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? 4.00 kg Box 6.00 kg Box Adding to eliminate T and find a Forces +

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Pulling a Box (Part 2) +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? Now for Tension We could have used the other tension formula from Box 2 and obtained the same answer Solve for Acceleration

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Example A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is connected by a 1kg rope and they are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? 1.00kg

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Pulling a Box (Part 3) Free body Diagram +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? 1.00 kg Because the rope has mass, the two ends will experience different tensions

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Pulling a Box (Part 3) +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? 4.00 kg Box6.00 kg BoxUsing F=ma for the system to find a Forces

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Pulling a Box (Part 3) +y +x a A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? Now for T1Solve for Acceleration Now for T2

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Example A worker drags a 38.0 kg box along the floor by pulling on a rope attached to the box. The coefficient of friction between the floor and the box is u s =0.450 and u k = a) What are the force of friction and acceleration when the worker applies a horizontal force of 150N? b) What are the force of friction and acceleration when the worker applies a horizontal force of 190N? 38kg

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Solution (Free Body Diagram) a) What are the force of friction and acceleration of the worker applies a horizontal force of 150N? 38kg The force of gravity down The Normal force up The applied force of tension to the right Friction to the left +y +x

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Solution (Vector Components) 38kg +y +x Since the applied force by the worker is only 150N, the box will not move What are the force of friction and acceleration of the worker applies a horizontal force of 150N? To determine if the box will move, we must find the maximum static friction and compare it to the applied force.

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Solution (Vector Components) 38kg +y +x b) What are the force of friction and acceleration of the worker applies a horizontal force of 190N? Since the applied force is greater than 168N from part a), we will have an acceleration in the x direction. So we will apply Newton’s 2 nd Law in the horizontal direction. The acceleration of the box is m/s 2 [E] F K =(0.410)(372.4N)=153N

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Example A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes. 8 kg 5 kg 13 kg 30 N We can find the acceleration of the train by treating the three masses as one unit. Tension in rope T 1 T1T1 T1T1 F Tension in rope T 2 T2T2 T1T1 T2T2 or

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Example A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 30 0 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and u k =0.40? 30 0

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Solution (Free Body Diagram) 30 0 The force of gravity down The Normal force up The applied force of tension at 30 0 Friction to the left Tension broken down into components A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 30 0 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and u k =0.40?

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) Solution (Force Components) 30 0 A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 30 0 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and u k =0.40? Vertical Components Horizontal Components We will need F N, so solve for F N Since we have a constant velocity, acceleration is 0 +y +x

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Solution (Force Components) 30 0 A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 30 0 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and u k =0.40? Solve for F T +y +x

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Example A group of children toboggan down a hill with a 30 0 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.

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Solution (Free Body Diagram) A group of children toboggan down a hill with a 30 0 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y Choose axis orientation to match the direction of motion and the normal to the surface Object Force of gravity is straight down Normal is perpendicular to the surface Force of friction opposes direction of motion Decompose gravity into axis components

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Solution (Force Vectors) A group of children toboggan down a hill with a 30 0 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y y direction Remember to solve for F N because we will need it later x direction

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+x +y Acceleration Example 9: Solution (Force Vectors) A group of children toboggan down a hill with a 30 0 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. Speed

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Example Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

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Solution (Free Body Diagram) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. Box B Object B Force of Gravity from A and B Normal Force Applied Force Friction from Table Tension Friction from A +y +x a

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Solution (Free Body Diagram) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. Box A Object Force of Gravity from A only Normal Force Tension Friction +y +x a +y +x a

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Solution (Force Vectors) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a A B x-direction for Box A x-direction for Box B This was determined using Box A

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Solution (Force Vectors) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a A B

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Example How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 kg 1.0 kg

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Example (free body diagram) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 kg 1.0 kg 3 Blocks taken as a Single Unit Object Force of Gravity Normal Force Applied Friction a +y +x 1.5 kg 2.0 kg 1.0 kg

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Example (force vectors) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 kg 1.0 kg a +y +x

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Example (free body diagram) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 1.5 kg 2.0 kg 1.0 kg 2.0 kg Block and 1.0 kg taken as a Single Unit Object Force of Gravity Normal Force Applied Friction a +y +x Since we are considering 2.0kg and 1.0kg block as a unit, then the Force is the push of 1.5 kg block on the combined block

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Example (force vectors) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 2.0 kg 1.0 kg a +y +x

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A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. a)Find the acceleration of the two objects b)Find the tension in the string. Step 1: Free body Diagram F G =9 kg FTFT FTFT The easiest way to choose the signs for the forces is to logical choose what you believe will be correct direction and follow that direction from one object to the other. If your final answer is negative, it just means your initial direction choice was wrong. + + Example m1m1

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A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. a)Find the acceleration of the two objects F G =9 kg FTFT FTFT + + Example (Solution) Horizonta Horizontal Vertical Adding these equations will remove the force of tension, thus giving allowing us to solve for acceleration. Since the answer is positive our initial direction choice was correct. m1m1 Note: don’t just say that a force of (9.8x9) is pulling on the 5 kg mass. This will give an acceleration of 17.6, which is greater than the acceleration due to gravity of the falling 9 kg mass, and you would not have tension in the rope. Ugly!

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A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object. b)Find the tension in the string. FGFG FTFT FTFT + + Example (Solution) Horizonta Horizontal We need only substitute the acceleration value into either the horizontal or vertical equation. Vertical m1m1

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Example Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block C descends with constant velocity. a)Determine the tension in the rope connecting Block A and B. b)What is the weight of Block C? c)If the rope connecting A and B were cut, what would be the acceleration of C?

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Example Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. a)Determine the tension in the rope connecting Block A and B. b)What is the weight of Block C? c)If the rope connecting A and B were cut, what would be the acceleration of C? Block A Object Gravity Normal Tension Block B Gravity Normal Tension from C Object Tension from A +x +y Friction

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Example (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. a)Determine the tension in the rope connecting Block A and B. b)What is the weight of Block C? c)If the rope connecting A and B were cut, what would be the acceleration of C? Block A Block B Block C

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Example (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. a)Determine the tension in the rope connecting Block A and B. b)What is the weight of Block C? c)If the rope connecting A and B were cut, what would be the acceleration of C? Block A Block B Block C Applying Block C’s equation to Block B Solving for the tension between block A and B The weight of Block C

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Example (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity. a)Determine the tension in the rope connecting Block A and B. b)What is the weight of Block C? c)If the rope connecting A and B were cut, what would be the acceleration of C? When the rope is cut:

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Example A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

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Example (Free Body Diagram) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. Object Gravity Tension Tension broken into components We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference.

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Example (Force Vectors) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. Vertical ForcesHorizontal Forces

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Example Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 30 0 to the horizontal. 5.0 kg 25N

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Example (Free Body Diagram) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 30 0 to the horizontal. 5.0 kg 25N Object Gravity Magnetic Applied Normal Since the gravity force down (5x9.8) is greater than force up (25sin(30), the box slides down, so friction is up. Friction

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Example (Vector Forces) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 30 0 to the horizontal. 5.0 kg 25N Horizontal Vertical +y +x

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Example (Insert Numbers) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 30 0 to the horizontal. 5.0 kg 25N +y +x

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Example Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is 0.15.

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Example (Free Body Diagram) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is For Masses 121 to 300 Big Mass Gravity Normal Tension from above Friction Treat objects 300 thru 121 as a single mass of 180*0.5kg

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Example (Vector Forces) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is For Combined Masses 121 to 300 +y +x a y-axis x-axis

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Example (Insert Values) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is For Combined Masses 1 to 120 +y +x a

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Example A person exerts a force of 175N at W30 0 S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a)Find the normal force on the crate b)Find the force of friction on the crate c)Find the acceleration of the crate a

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Example (Free Body Diagram) A person exerts a force of 175N at W30 0 S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a)Find the normal force on the crate b)Find the force of friction on the crate c)Find the acceleration of the crate a FgFg FNFN FfFf FaFa +y +x

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Example (Vector Forces) A person exerts a force of 175N at W30 0 S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a)Find the normal force on the crate b)Find the force of friction on the crate c)Find the acceleration of the crate. FgFg FNFN FfFf FaFa 30 0 a y-axis Insert Values +y +x

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Example (Vector Forces) A person exerts a force of 175N at W30 0 S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a)Find the normal force on the crate b)Find the force of friction on the crate c)Find the acceleration of the crate. FgFg FNFN FfFf FaFa 30 0 a Friction +y +x

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Example (Vector Forces) A person exerts a force of 175N at W30 0 S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a)Find the normal force on the crate b)Find the force of friction on the crate c)Find the acceleration of the crate. FgFg FNFN FfFf FaFa 30 0 a Acceleration +y +x

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Example Calculate the unknowns for each accelerated block. 18 kg a) F 6 kg b) a m m c)

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Example (Solution) Calculate the unknowns for each accelerated block. 18 kg a) F

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Example (Solution) Calculate the unknowns for each accelerated block. 6 kg b) a

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Example (Solution) Calculate the unknowns for each accelerated block. m m c)

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Example Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μ s

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Example Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μ s If I do not fall, then the friction force, F f, must balance my weight mg, that is F f = mg The only horizontal force is the Normal Force.Therefore F=N=ma Putting this together we obtain:

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Attached bodies on two inclined planes all surfaces frictionless peg is frictionless m1m1 m2m2 smooth peg 11 22

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How will the bodies move? From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law: Taking “x” components: 1) T 1 - m 1 g sin 1 = m 1 a 1X 2) T 2 - m 2 g sin 2 = m 2 a 2X But T 1 = T 2 = T and -a 1X = a 2X = a (constraints) m1m1 y x m2m2 x y T1T1 N m1gm1g 11 m2gm2g T2T2 N 22

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Solving the equations Using the constraints, we get 2 eqn and 2 unks, solve the equations. T - m 1 gsin 1 = -m 1 a(a) T - m 2 gsin 2 = m 2 a(b) Subtracting (a) from (b) gives: m 1 gsin 1 - m 2 gsin 2 = (m 1 +m 2 )a So: a mm mm g sin

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Two-body dynamics In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. (a) (b) (c) (a) Case (1) (b) Case (2) (c) same m 10kg aa m F = 98.1 N Case (1)Case (2)

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Solution m 10kg a l Add (a) and (b): m W g = (m + m W )a l Note: (a) (b) T = ma (a) m W g -T = m W a (b) l For case (1) draw FBD and write F NET = ma for each block: m W =10kg

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Solution l The answer is (b) Case (2) In this case the block experiences a larger acceleratioin T = 98.1 N = ma l For case (2) m 10kg a Case (1) m a F = 98.1 N Case (2)

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Problem: Accelerometer A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g. A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g. a i

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Accelerometer... Draw a free body diagram for the mass: Draw a free body diagram for the mass: What are all of the forces acting? What are all of the forces acting? m T T (string tension) g mg (gravitational force) i

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Accelerometer... Using components (recommended): Using components (recommended): i: F X = T X = T sin = ma i: F X = T X = T sin = ma j: F Y = T Y mg j: F Y = T Y mg = T cos mg = 0 = T cos mg = 0 T gmggmg m amaama ji TXTX TYTY

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Accelerometer... Using components : Using components : i: T sin = ma i: T sin = ma j: T cos - mg = 0 j: T cos - mg = 0 Eliminate T : Eliminate T : gmggmg m amaama T sin = ma T cos = mg TXTX TYTY ji T

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Accelerometer... Alternative solution using vectors (elegant but not as systematic): Alternative solution using vectors (elegant but not as systematic): Find the total vector force F NET : Find the total vector force F NET : T gmggmg F F TOT m T T (string tension) g mg (gravitational force)

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Accelerometer... Alternative solution using vectors (elegant but not as systematic): Alternative solution using vectors (elegant but not as systematic): Find the total vector force F NET : Find the total vector force F NET : Recall that F NET = ma: Recall that F NET = ma: So So amaama T gmggmg m T T (string tension) g mg (gravitational force)

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Accelerometer... Let’s put in some numbers: Let’s put in some numbers: Say the car goes from 0 to 60 mph in 10 seconds: Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s. 60 mph = 60 x 0.45 m/s = 27 m/s. Acceleration a = Δv/Δt = 2.7 m/s 2. Acceleration a = Δv/Δt = 2.7 m/s 2. So a/g = 2.7 / 9.8 = So a/g = 2.7 / 9.8 = = arctan (a/g) = 15.6 deg = arctan (a/g) = 15.6 deg a

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Understanding A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces (A)Have equal magnitudes and form an action/reaction pair (B)Have equal magnitudes but do not form an action/reaction pair (C)Have unequal magnitudes and form an action/reaction pair (D)Have unequal magnitudes and do not form an action/reaction pair (E)None of the above Because the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

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