A series of blocks is connected to a pulley in the manner depicted below. Find the minimum mass that block C must have in order to keep block A

from sliding on the table top.

ms A

C

B

mA=4 KgmB=2 Kg

ms=0.2 (coefficient of static friction)

The pulley and rope are to be considered massless.y

x

Question 1Question 2Question 3Question 4Question 5Question 6Question 7Reflection

1. Which physics principle should we use to solve this problem.

A) Work-Energy Theorem

B) Newton’s 2nd Law

C) Law of conservation of energy

Choice AIncorrect

Since the blocks are not in motion, there is no work being done. The work-energy

theorem will not help us solve this problem.

Since we are considering a situation where the blocks are not slipping on the table and the hanging block is not moving up or

down, the system is in static equilibrium.

Newton’s Laws tell us that the sum of the forces on an object or the entire system is zero for mechanical equilibrium. In component

form in two dimensions, the conditions are:

Choice BCorrect

Fx 0 and Fy 0

The law of conservation of energy is of no use here.

Try again. Remember that none of the blocks are in motion in our situation.

Choice CIncorrect

2. Let’s begin by analyzing the hanging block B.

Which of the following free body diagrams correctly depicts all of the forces acting on block B?

A)

B)

C)

mBg

T

mBg

mAg+mCg

T

mBg

mAg+mCg

T= tensiong=9.8m/s2

There are no forces acting on the hanging block in the positive or negative

x-direction.

This diagram correctly shows the only two forces acting on the block.

Choice ACorrect

The weight of blocks A and C do not act on the hanging block directly. They pull on the rope

that is connected to block B.

The only forces that act on block B are the tension and the block’s weight.

Choice BIncorrect

Choice CIncorrect

There are no forces acting on the hanging block in the positive or negative

x-direction.

3. Using the free body diagram for block B, and applying Newton’s 2nd Law we get which

one of the following relations?

A)

B)

C)

T mBg

T mBg

mAgmCg mBg

Choice ACorrect

Fy T mB g 0

T mBg

Using Newton’s 2nd Law and our free body diagram we find:

After writing down Newton’s 2nd Law resolved into x and y components, we only find a useful equation from the y-component, because none of the

forces have x-components.

Choice BIncorrect

Using Newton’s 2nd Law and our free body diagram we find:

Fy T mB g 0

T mBg The two forces are subtracted because they act in opposite directions.

Check the free body diagram that we found to be correct in the previous

question and try again.

Choice CIncorrect

4. Let’s now analyze blocks A and C.

Which of the following free body diagrams correctly depicts all of the forces acting on blocks A and C if we treat them as one

system?

A)

B)

C)

TFf

(mA-mC)g

N

N

N

FfT(mA+mC)g

Ff T

(mA+mC)g

Ff=frictional forceN=normal force

The directions of the forces of friction and tension should be switched. The force of

static friction opposes the sliding which would be to the right if block C were not massive

enough.

The rope is attached to block A on the right side, so the tension would act away from the

block in the positive x-direction.

Choice AIncorrect

This diagram correctly depicts all of the forces acting on the system of

blocks A and C.

Choice BCorrect

The combined weight W of blocks A and C should be considered. We are

treating them together as one system.

Choice CIncorrect

W=(mA-mC)g W=(mA+mC)g

5. Use the free body diagram from the previous question.

Which of the following relations do we get by applying Newton’s 2nd Law to blocks A & C together as one system?

(Resolve the forces into x and y components.)

A)

B)

C)

T Ff

N mA mC g

Fx 0

Fy 0

T Ff

T mA mC g

N mA mC g

N mA mC g

Choice ACorrect

Fx T Ff 0

T Ff

Fy N (mA mC)g 0

N(mA mC)g

Using Newton’s 2nd Law and our free body diagram we find:

Choice BIncorrect

Using Newton’s 2nd Law and our free body diagram we find:

Fy N (mA mC)g 0

N(mA mC)g

Fx T Ff 0

T Ff

The weight of the system of blocks A and C has no x-component.

Choice CIncorrect

6. Use the information from the last question, to find another way to express T in terms of known quantities and the quantity that we are interested in finding, which is the minimum mass (mC) that block C can have to keep the

system in equilibrium.

Which one of the following expressions is correct?

A)

B)

C)

Note:The force of static friction that

will keep the blocks from slipping must satisfy the

following condition:

We want to find the minimum mass of block C that allows this condition to be satisfied. This occurs when:

Ff msN

T ms mA mC g

T ms mA mB mC g

T mA mB mC g

Ff msN

Choice AIncorrect

N mA mC gFf msNms mA mC gT Ff

T ms mA mC g

Combine several equations that we have

found in previous questions,

Choice BIncorrect

Combine several equations that we have

found in previous questions.

N mA mC gFf msNms mA mC gT Ff

T ms mA mC g

Choice CCorrect

We combine several equations that we have

found in previous questions.

N mA mC gFf msNms mA mC gT Ff

T ms mA mC g

7. We now have two equations for the tension in the rope T. If we equate the two, we can solve

for the minimum mass of block C.

What is the minimum mass that block C must have in order to keep block A from sliding on the

table top?

A) mC=10kg

B) mC=6kg

C) mC=4kg

Choice AIncorrect

mC mB

ms

mA

Please try again. You should find the following expression for the

minimum mass of block C:

Begin by equating these two expressions for tension.

T mBgT ms mA mC g

Choice BCorrect

T mBgT ms mA mC g

mBg ms mA mC gmB ms mA mC

mB

ms

mA mC

mB

ms

mA mC

mC mB

ms

mA

mC 2kg0.2

4kg 6kg

mC 6kg

The force of friction on the combined mass (mA+mC) is balanced by the tension force. Block C must have a mass of at least 6 kg in order to keep the other blocks from moving.

Choice CIncorrect

Please try again. You should find the following expression for the

minimum mass of block C:

mC mB

ms

mA

Begin by equating these two expressions for tension.

T mBgT ms mA mC g

Reflection Questions

• If the coefficient of static friction was lower than 0.2, what would happen to blocks A and C?

• If the coefficient of static friction was higher than 0.2, what would happen to blocks A and C? More

Reflection Question:

• How do we know that the relation:

gives the minimum mass required to keep the blocks from slipping?

Ff msN

Help

T mBgT ms mA mC g

mBg ms mA mC gmB ms mA mC

mB

ms

mA mC

mB

ms

mA mC

Using the inequality we find:

This means that the mass of block C must be greater than or equal to

6 kg in order to keep the blocks from

slipping. Obviously, 6kg is the

minimum mass to satisfy this condition, so if we just use the equal sign originally, we find

the minimum mass.

2kg0.2

4kg mC

6kg mC