EXAMPLE:



NEWTON’S LESSON 15 COUPLED SYSTEMS I

EXAMPLE:

In figure a), a constant horizontal force, Fapp of magnitude 20.0 N is applied to block A of mass mA = 4.0 kg, which pushes against block B of mass mB = 6.0 kg. The blocks slide over a frictionless surface, along an x axis. (a) What is the acceleration of the blocks?

(b) What is the force FBA on block B from A?

i. Draw a labeled free-body diagram for the two boxes during the time when the boxes are accelerating, specifying all the forces acting on the block. (Be sure to specify the type of force and the object causing each force.) Wherever you can, compare the magnitudes of forces.

Recall action-reaction forces : box A must be pushing on box B with the same force as box B is pushing on box A.

ii. Write equations for the free body diagrams for box A and box B:

Box B: FnetB = FBA

Box A: FnetA = Fapp – FAB

iii.

NOTE:

The force net of each box is the total sum force on the box which is its mass times its acceleration.

FnetB = mAaA

FnetA = mBaB

The boxes must have identical acceleration otherwise the boxes would not be connected. aB = aA

Unfortunately, we do not know each Fnet, therefore we must substitute into the other equation to find the acceleration of box A and B. Note: the acceleration of the boxes is equal.

FnetA = Fapp – FAB

substituting for FAB (FAB is = but directionally opp to FBA)

FnetA = Fapp – FnetB

mA a = Fapp - mB a

Fapp = (mA+ mB)a

a = Fapp/(mA+mB) = 20.0/(4.0 + 6.0) = 2.0 m/s2

b. What is the force FBA on block B from block A?

i. FBA = mBa = 6.0*2.0 = 12 N The acceleration on block A and B are the same.

EXAMPLE: TWO BOXES AND A CORD

Two boxes resting on a frictionless surface are connected by a massless cord. The boxes have masses of 12.0 and 10.0 kg. A horizontal force of Fapp = 40.0 N is applied by pulling on the lighter box.

What is the acceleration of each box and the tension in the cord?

[pic]

i. Draw a freebody diagram for each box:

[pic]

ii. Write equations for the free body diagrams for box 1 and box 2:

Box 1: Fnet1 = Fapp – FT

Box 2: Fnet2 = FT

iii. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch up which is counter to our experience of tension between the boxes.

In order to find the acceleration, we must add the two equations:

Fnet1 = Fapp – Fnet2

m1a = Fapp – m2a

Fapp = a(m1 + m2)

a = 40.0/22.0 = 1.82 m/s2

To determine tension, substitute m2 into box 2 equation (could use box 1 equation as well): Remember that the force of tension is the pull of box 2 on box 1, therefore it is dependent on the mass at the end of the string.

FT = m2a = 12.0 * 1.82 = 21.8 N

It makes sense for the tension to be less than the pulling force since the tension accelerates only the second box.

Note how analyzing two free-body diagrams allows us to understand an “internal force” like Tension in the rope. Summed together, tension would make zero contribution to the net force on the whole system.

EXAMPLE: Stacked Boxes

Block A is attached to the wall and is sitting on Block B. If block A is stationary and Block B doesn’t accelerate as it is being pulled left, out from underneath Block A, what is the Force applied to Block B and the Tension in the rope attached to Block A. The coefficient of friction between the blocks and Block B and the surface on which it slides is 0.20.

[pic]

i. First draw a free body diagrams for both blocks:

[pic]

ii. Write equations for the free body diagrams for block 1 and block 2:

Block 1: Fnet1 = FT - Ff

Block 2: Fnet2 = Fapp - Ff- Ff

iii. If both boxes have zero acceleration, determine the force applied and force tension using a = 0:

Block 1: Fnet1 = FT - Ff

m1a1 = FT - μmg

0 = FT – 0.20*2.5*9.8

FT = 4.9 N

Block 2: Fnet2 = Fapp - Ff - Ff

m2a2 = Fapp - μm1g - μ(m1+m2)g

0 = Fapp – 0.2*2.5*9.8 – 0.2*7.5*9.8

Fapp = 4.9 + 14.7 = 19.6 N

NEWTON’S LESSON 15 HOMEWORK

1. A 1.3 N force is applied to the coupled system that has no friction. The system accelerates at 1.85 m/s2. Find m2.

[pic]

2. Find the acceleration of the system and tension in the connecting rope:

[pic]

3. Find the acceleration of the system and the tension in the connecting rope:

[pic]

4. Two crates, of mass 75 kg (box A) and 110 kg (box B), are in contact and at rest on a horizontal surface. A 620-N force is exerted on the 75-kg crate. If the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other. (c) Repeat with the crates reversed (no answer key)

5. Block A (mass 3.4 kg) is attached to the wall and is sitting on Block B (mass 6.0 kg). If block A is stationary and Block B doesn’t accelerate as it is being pulled left, out from underneath Block A, what is the Force applied to Block B and the Tension in the rope attached to Block A. The coefficient of friction between the blocks and Block B and the surface on which it slides is 0.32.

HOMEWORK KEY

1. 0.25 kg, 0.47 N

2. 2.0 m/s2, 7.0 N

3. 3.0 m/s2, 0.90 N, 2.1 N

4. 4.8 m/s2, 530 N

5. 11 N, 29 N

NEWTON’S LESSON 15 HOMEWORK

1. A 1.3 N force is applied to the coupled system that has no friction. The system accelerates at 1.85 m/s2. Find m1.

[pic]

i. Draw a freebody diagram for each box:

[pic]

ii. Write equations for the free body diagrams for box 1 and box 2:

Box 1: Fnet1 = Fapp – FT

Box 2: Fnet2 = FT

iii. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch up which is counter to our experience of tension between the boxes.

In order to find the acceleration, we must add the two equations:

Fnet1 = Fapp – Fnet2

m1a = Fapp – m2a

0.45(1.86) = 1.3 – m2(1.86)

(1.3 - 0.837)/1.86 = m2 0.2489 --> 0.25 kg

m2 = 0.2489 = 0.25 kg

To determine tension, substitute m2 into box 2 equation (could use box 1 equation as well):

Fnet2 = FT

m2a = FT

FT = 0.47 N

2. Find the acceleration of the system and tension in the connecting rope:

[pic]

i. Draw a freebody diagram for each box:

[pic]

ii. Write equations for the free body diagrams for box 1 and box 2:

Box 1: Fnet1 = Fapp – FT

Box 2: Fnet2 = FT

iii. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch up which is counter to our experience of tension between the boxes.

In order to find the acceleration, we must add the two equations:

Fnet1 = Fapp – Fnet2

m1a = Fapp – m2a

Fapp = a(m1 + m2)

a = 12/6.0 = 2.0 m/s2

Force of tension is determined by substituting a into box 1 equation:

Fnet1 = Fapp – FT

Fapp - FT = ma

12 – FT = 2.5(2.0)

FT = 7.0 N

3. Find the acceleration of the system and the tension in teh connecting rope:

[pic]

i. Draw a freebody diagram for each box:

[pic]

ii. Write equations for the free body diagrams for box 1, box 2 and box 3:

Box 1: Fnet1 = FT1

Box 2: Fnet2 = FT2 – FT1

Box 3: Fnet3 = Fapp – FT2

iii. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch up which is counter to our experience of tension between the boxes.

In order to find the acceleration, we must substitute the three equations:

Fnet3 = Fapp – FT2

Fnet3 = Fapp – Fnet2 – FT1

Fnet3 = Fapp – Fnet2 – Fnet1

m3a = Fapp – m2a – m1a

a(m3 + m2 + m1) = Fapp

3.6 = a(0.30 + 0.40 + 0.50)

a = 3.0 m/s2

In order to find tension, substitute a into the following equations:

Box 1: Fnet1 = FT1

m1a = FT1

0.30(3.0) = 0.90 N

Box 2: Fnet2 = FT2 – FT1

m2a = FT2 – 0.90

0.40(3.0) + 0.90 = FT2

FT2 = 2.1 N

4. Two crates, of mass 75 kg (box A) and 110 kg (box B), are in contact and at rest on a horizontal surface. A 620-N force is exerted on the 75-kg crate. If the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other. (c) Repeat with the crates reversed.

i. Draw a labeled free-body diagram for the two boxes during the time when the boxes are accelerating, specifying all the forces acting on the block. (Be sure to specify the type of force and the object causing each force.) Wherever you can, compare the magnitudes of forces.

Recall action-reaction forces : box A must be pushing on box B with the same force as box B is pushing on box A.

ii. Write equations for the free body diagrams for box A and box B:

Box B: FnetB = FBA - FfB

Box A: FnetA = Fapp – FAB - FfA

NOTE:

The force net of each box is the total sum force on the box which is its mass times its acceleration.

FnetB = mAaA

FnetA = mBaB

The boxes must have identical acceleration otherwise the boxes would not be connected. aB = aA

Unfortunately, we do not know each Fnet, therefore we must substitute into the other equation to find the acceleration of box A and B. Note: the acceleration of the boxes is equal.

FnetA = Fapp – FAB - FfA

substituting for FAB (FAB is = but directionally opp to FBA)

FnetA = Fapp – (FnetB + FfB) - FfA

mA a = Fapp - mB a -μmBg - μmAg

Fapp = (mA+ mB)a – 0.15(9.8)( mA+ mB)

a = (Fapp + 0.15(9.8)( mA+ mB))/(mA+mB) = 4.821 = 4.8 m/s2

c. What is the force FBA on block B from block A (each exerts the same force on the other, definition of action-reaction)?

ii. FBA = mBa = 110*4.821 = 530 N

5. Block A (mass 3.4 kg) is attached to the wall and is sitting on Block B (mass 6.0 kg). If block A is stationary and Block B doesn’t accelerate as it is being pulled left, out from underneath Block A, what is the Force applied to Block B and the Tension in the rope attached to Block A. The coefficient of friction between the blocks and Block B and the surface on which it slides is 0.32.

i. First draw a free body diagrams for both blocks:

ii. Write equations for the free body diagrams for block 1 and block 2:

Block 1: Fnet1 = FT - Ff

Block 2: Fnet2 = Fapp - Ff- Ff

iii. If both boxes have zero acceleration, determine the force applied and force tension using a = 0:

Block 1: Fnet1 = FT - Ff

m1a1 = FT - μmg

0 = FT – 0.32*3.4*9.8

FT = 10.6624 = 11 N

Block 2: Fnet2 = Fapp - Ff - Ff

m2a2 = Fapp - μm1g - μ(m1+m2)g

0 = Fapp – 0.32*3.4*9.8 – 0.32*6.0*9.8

Fapp = 10.6624 + 18.816 = 29 N

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