Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the.

Force Problems

A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the car in the direction of motion is (a) 0 N (b) Greater than 2000 N (c) Exactly equal to 2000 N (d) Less than 2000 N. (e) answers (a) and (d) are correct

The unit Newton (N) is equivalent to (a) kg m/s 2 (b) kg m/s (c) m/s 2 (d) kg m 2 /s 2 (e) kg m/s

A 5.0 kg object experiences an acceleration of 2.0 m/s 2. If the mass was halved and the net force doubled, the object would (a) move at a constant velocity. (b) would accelerate at 8.0 m/s 2 (c) would accelerate at 2.0 m/s 2 (d) would accelerate at 1.0 m/s 2 (e) would accelerate at 0.5 m/s 2

The velocity-time graph for an object is a non-zero, horizontal line. Which of the following situations might this describe? (a) The object experiences a net force of 0 N. (b) The object is at rest. (c) The object is accelerating. (d) There are no forces acting on the object. (e) answers (c) and (d) are correct.

Assuming an object with m = 2 kg, what kind of motion results from the following force diagram,? (a) constant velocity (b) accelerates 4.0 m/s 2 [right] (c) accelerates 5.0 m/s 2 [right] (d) accelerates 3.0 m/s 2 [right] (e) Accelerates 5.0 m/s 2 [left]

Example 1 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 If F T was greater than F g then the 5kg mass would go up. If F T was less than F g then the 5kg was would go down.

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?

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

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

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?

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?

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 +

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

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

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

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

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

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

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 =0.410. 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

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

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.

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 0.982 m/s 2 [E] F K =(0.410)(372.4N)=153N

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

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?

) 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

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

Example The tension in the horizontal rope is 30N A) Determine the weight of the object 30N 40 0 50 0 Diagram Free body Diagram 50 0

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

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. 10 0 15 0 80N T1T1 T2T2 Diagram Free body Diagram

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

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

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:

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

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

-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

-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

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

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.

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

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

+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

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

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

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

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

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

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

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

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. F G =9 kg FTFT FTFT + + Example (Solution) Horizonta Horizontal We need only substitute the acceleration value into either the horizontal or vertical equation. Vertical m1m1

Example A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.2 0 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

Example (Free Body Diagram) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.2 0 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.

Example (Force Vectors) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.2 0 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. Vertical ForcesHorizontal Forces

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

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

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

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

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. 30 0 a

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. 30 0 a FgFg FNFN FfFf FaFa +y +x

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

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

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

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

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

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

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

Example Suppose the coefficient of Kinetic Friction between m 1 and the ramp is u k =0.15, and both masses are 3.0 kg. a) If m 2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of u k, that will keep the system from accelerating?

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is u k =0.15, and both masses are 3.0 kg. a) If m 2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of u k, that will keep the system from accelerating? m2m2 F g =m 2 g FTFT m1m1 FNFN F g =m 1 g FTFT

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is u k =0.15, and both masses are 3.0 kg. a) If m 2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of u k, that will keep the system from accelerating? m2m2 F g =m 2 g FTFT m1m1 FNFN F g =m 1 g Because the two masses are connected, we can treat them as one unit and just apply the forces that move it one way or another.

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is u k =0.15, and both masses are 3.0 kg. a) If m 2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of u k, that will keep the system from accelerating? m2m2 F g =m 2 g FTFT m1m1 FNFN F g =m 1 g

m1m1 y x m2m2 x y T1T1 N m1gm1g 11 m2gm2g T2T2 N 22 Attached bodies on two inclined planes Free Body Diagram Step 1 Pick a direction in which you think the blocks will move and make that direction positive +

We want to eliminate T, so let’s add Since the acceleration is negative, our original choice for positive direction was wrong, so mass 1 goes up, and mass 2 goes down both at 2.2 m/s 2. Block 1 Block 2 m1m1 y x m2m2 x y T1T1 N m1gm1g 11 m2gm2g T2T2 N 22 +

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)

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

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)

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.

Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the.

Similar presentations

Presentation on theme: "Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the.

Similar presentations

Presentation on theme: "Force Problems. A car is traveling at constant velocity with a frictional force of 2000 N acting opposite the motion of the car. The force acting on the."— Presentation transcript:

Similar presentations

About project

Feedback