Force Problems.

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/s2 (b) kg m/s (c) m/s2 (d) kg m2/s2 (e) kg m/s

A 5. 0 kg object experiences an acceleration of 2. 0 m/s2
A 5.0 kg object experiences an acceleration of 2.0 m/s2. 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/s2 (c) would accelerate at 2.0 m/s2 (d) would accelerate at 1.0 m/s2 (e) would accelerate at 0.5 m/s2

The velocity-time graph for an object is a non-zero, horizontal line
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/s2 [right] (c) accelerates 5.0 m/s2 [right] (d) accelerates 3.0 m/s2 [right] (e) Accelerates 5.0 m/s2 [left]

The gravitational field strength on the surface of the Moon is 1
The gravitational field strength on the surface of the Moon is 1.6 N/kg. What would a person weigh on Mars if this person weighs 500 N on Earth? We first need to determine the person’s Mass Now for the Weight on the Moon.

Example 1 The system below is in equilibrium. If the scale is calibrated in N, what does it read? Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N If FT was greater than Fg then the 5kg mass would go up. If FT was less than Fg then the 5kg was would go down. 5kg 5kg

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? +y +x a Vertical Forces Horizontal Forces Solving

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) 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? Free body Diagram +y +x a

Pulling a Box (Part 2) + Forces
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? a +y +x Forces 4.00 kg Box 6.00 kg Box Adding to eliminate T and find a +

Pulling a Box (Part 2) 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? a +y +x Solve for Acceleration Now for Tension We could have used the other tension formula from Box 2 and obtained the same answer

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) 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? Free body Diagram 1.00 kg Because the rope has mass, the two ends will experience different tensions +y +x a

Pulling a Box (Part 3) Forces
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? a +y +x Forces 4.00 kg Box 6.00 kg Box Using F=ma for the system to find a

Pulling a Box (Part 3) 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? a +y +x Solve for Acceleration Now for T1 Now for T2

Example A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes. T1 T2 8 kg 5 kg 13 kg 30 N We can find the acceleration of the train by treating the three masses as one unit. T2 T1 Tension in rope T2 or T1 F Tension in rope T1

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 us=0.450 and uk=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? The Normal force up Friction to the left The applied force of tension to the right 38kg The force of gravity down +y +x

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

Solution (Vector Components)
b) What are the force of friction and acceleration of the worker applies a horizontal force of 190N? 38kg +y +x 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 2nd Law in the horizontal direction. FK=(0.410)(372.4N)=153N The acceleration of the box is m/s2 [E]

Example A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 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 uk=0.40? 300

A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 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 uk=0.40? The Normal force up Friction to the left The applied force of tension at 300 300 Tension broken down into components The force of gravity down

Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 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 uk=0.40? 300 +y +x Vertical Components Horizontal Components Since we have a constant velocity, acceleration is 0 We will need FN, so solve for FN

Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 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 uk=0.40? 300 +y +x Solve for FT

Example Diagram Free body Diagram
The tension in the horizontal rope is 30N A) Determine the weight of the object Diagram Free body Diagram 30N 400 500 500

A) Determine the weight of the object
500 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 Diagram Free body 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. Diagram Free body Diagram 100 150 80N T1 T2

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 Free body Diagram Block Ring
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
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. Ring Block From Block: From ring: Combining:

Atwood’s Machine Example:
Masses m1 = 10 kg and m2 = 20kg are attached to an ideal massless string and hung as shown around an ideal massless pulley. What are the tensions in the string T1 and T2 ? Find the accelerations, a1 and a2, of the masses. Fixed Pulley T1 T2 m1 a1 m2 a2

Draw free body diagrams for each object
Applying Newton’s Second Law: T1 - m1g = m1a (a) T2 - m2g = -m2a2 => -T2 + m2g = +m2a2 (b) But T1 = T2 = T since pulley is ideal and a1 = -a2 =a since the masses are connected by the string m2g m1g Free Body Diagrams T1 T2 a1 a2

Solve for Acceleration
-m1g + T = m1 a (a) -T + m2g = m2 a (b) Two equations and two unknowns we can solve for both unknowns (T and a). Add (b) + (a): g(m2 – m1 ) = a(m1+ m2 ) m2g m1g Free Body Diagrams T1 T2 a1 a2

Solve for T -m1g + T = m1 a (a) -T + m2g = m2 a (b)
Plug a into (b) and Solve for T m2g m1g Free Body Diagrams T1 T2 a1 a2

Atwood Machine Review So we find: m1 m2 a T

Example A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.

A group of children toboggan down a hill with a 300 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 Normal is perpendicular to the surface Object Decompose gravity into axis components Force of friction opposes direction of motion Force of gravity is straight down

Remember to solve for FN because we will need it later
Solution (Force Vectors) A group of children toboggan down a hill with a 300 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 x direction Remember to solve for FN because we will need it later

Example 9: Solution (Force Vectors)
A group of children toboggan down a hill with a 300 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 Acceleration Speed

Example How much force is needed to give the blocks an acceleration of 3.0 m/s2 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 1.0

Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 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? 3 Blocks taken as a Single Unit Normal Force 1.5 kg 2.0 1.0 1.5 kg 2.0 1.0 Friction Object Applied +y +x Force of Gravity a

Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 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 1.0 +y +x a

Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 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 Block and 1.0 kg taken as a Single Unit 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 Normal Force Friction Object 2.0 kg Applied 1.5 kg 1.0 kg Force of Gravity +y +x a

Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 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 +y +x a

Example Step 1: Free body Diagram FT FT m1 + FG=9 kg +
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. Find the acceleration of the two objects Find the tension in the string. Step 1: Free body Diagram FG=9 kg FT FT m1 + + 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 (Solution) FT FT m1 + FG=9 kg +
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. Find the acceleration of the two objects FG=9 kg FT FT m1 + + Adding these equations will remove the force of tension, thus giving allowing us to solve for acceleration. Horizontal Vertical Since the answer is positive our initial direction choice was correct.

Example (Solution) FT FT m1 + FG=9 kg +
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. Find the tension in the string. FG=9 kg FT FT m1 + + We need only substitute the acceleration value into either the horizontal or vertical equation. Horizontal Vertical

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.

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. We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference. Tension broken into components Tension Object Gravity

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 Forces Horizontal 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 300 to the horizontal. 5.0 kg 25N

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 300 to the horizontal. Since the gravity force down (5x9.8) is greater than force up (25sin(30), the box slides down, so friction is up. Applied Friction 5.0 kg 25N Object Magnetic Normal Gravity

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 300 to the horizontal. 5.0 kg 25N +y +x Vertical Horizontal

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 300 to the horizontal. 5.0 kg 25N +y +x

Example A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. 300 a

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. 300 FN a Ff 300 +y +x Fa Fg

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. +y +x Fg FN Ff Fa 300 a y-axis Insert Values

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. Fg FN Ff Fa 300 +y +x a Friction

A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, Find the normal force on the crate Find the force of friction on the crate Find the acceleration of the crate. Fg FN Ff Fa 300 +y +x a Acceleration

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

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

6 kg b) a

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating?

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 m1 FN FT FT Fg=m2g Fg=m1g

Example Suppose the coefficient of Kinetic Friction between m1 and the ramp is uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 Fg=m2g FT m1 FN Fg=m1g 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 uk=0.15, and both masses are 3.0 kg. a) If m2 moves down, determine the magnitude of the acceleration of the masses. b) What is the smallest value of uk, that will keep the system from accelerating? m2 Fg=m2g FT m1 FN Fg=m1g

Two-body dynamics Case (1) Case (2)
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. m 10kg a a m F = 98.1 N Case (1) Case (2) (a) Case (1) (b) Case (2) (c) same

Solution For case (1) draw FBD and write FNET = ma for each block: (a) T = ma (a) mWg -T = mWa (b) m 10kg a Add (a) and (b): mWg = (m + mW)a mW=10kg (b) Note:

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

Example An elevator has a total mass of 2000 kg. What tension force must the cables provide to accelerate the elevator from rest to a velocity of 10.0 m/s [down] in 5.0 s? We need to find the acceleration, so we can solve the Force of Tension. Gravity Tension Now Let’s plug values into our F=ma equation and solve for Tension.

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 Have equal magnitudes and form an action/reaction pair Have equal magnitudes but do not form an action/reaction pair Have unequal magnitudes and form an action/reaction pair Have unequal magnitudes and do not form an action/reaction pair 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.

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