# Pulley Problems

On this page I put together a collection of pulley problems to help you understand pulley systems better. The required equations and background reading to solve these problems are given on the

friction page, the

equilibrium page, and

Newton's second law page.

**Problem # 1**
A block of mass

*m* is pulled, via pulley, at constant velocity along a surface inclined at angle

*θ*. The coefficient of kinetic friction is

*μ*_{k}, between block and surface. Determine the pulling force

*F*.

__Answer:__ *mg*cos

*θ**μ*_{k}+

*mg*sin

*θ*
**Problem # 2**
Two blocks of mass

*m* and

*M* are hanging off a single pulley, as shown. Determine the acceleration of the blocks. Ignore the mass of the pulley.

Hint and answer
**Problem # 3**
Two blocks of mass

*m* and

*M* are connected via pulley with a configuration as shown. The coefficient of static friction is

*μ*_{s}, between block and surface. What is the maximum mass

*m* so that no sliding occurs?

__Answer:__ maximum

*m* =

*M**μ*_{s}
**Problem # 4**
Two blocks of mass

*m* and

*M* are connected via pulley with a configuration as shown. The coefficient of static friction is

*μ*_{s}, between block and surface. What is the minimum and maximum mass

*M* so that no sliding occurs?

Hint and answer
**Problem # 5**
Two blocks of mass

*m* and

*M* are connected via pulley with a configuration as shown. The coefficient of static friction is

*μ*_{s}, between blocks and surface. What is the maximum mass

*m* so that no sliding occurs?

__Answer:__ Maximum

*m* =

*M**μ*_{s}/(sin

*θ*—cos

*θ**μ*_{s})

**Problem # 6**
Two blocks of mass

*m* and

*M* are connected via pulley with a configuration as shown. The coefficient of static friction between the left block and the surface is

*μ*_{s1}, and the coefficient of static friction between the right block and the surface is

*μ*_{s2}. Formulate a mathematical inequality for the condition that no sliding occurs. There may be more than one inequality.

Hint and answer
**Problem # 7**
A block of mass

*m* is pulled, via two pulleys as shown, at constant velocity along a surface inclined at angle

*θ*. The coefficient of kinetic friction is

*μ*_{k}, between block and surface. Determine the pulling force

*F*. Ignore the mass of the pulleys.

Hint and answer
**Problem # 8**
A block of mass

*m* is lifted at constant velocity, via two pulleys as shown. Determine the pulling force

*F*. Ignore the mass of the pulleys.

Hint and answer
**Problem # 9**
A block of mass

*M* is lifted at constant velocity, via an arrangement of pulleys as shown. Determine the pulling force

*F*. Ignore the mass of the pulleys.

Hint and answer
The hints and answers for these pulley problems will be given next.

**Hints And Answers For Pulley Problems**
__Hint and answer for Problem # 2__
This is called the Atwood machine and is commonly used for demonstration in physics classes.

Apply Newton's second law to the block on the left. We have

*Mg*—

*T* =

*Ma* (taking the downward direction as positive). Apply Newton's second law to the block on the right. We have

*mg*—

*T* = -

*ma* (the acceleration of the two blocks have opposite signs, since one moves up and the other moves down). Combine these two equations and we can find an expression for the acceleration of the blocks.

__Answer:__ *a* = (

*M*—

*m*)

*g*/(

*M*+

*m*)

__Hint and answer for Problem # 4__
For the maximum mass

*M*, the block is on the verge of sliding down the incline. This means that

*Mg*sin

*θ*—

*T*—

*Mg*cos

*θ**μ*_{s} = 0, where

*T* is the tension in the rope. Since

*T* =

*mg*, we can calculate the maximum

*M* from the previous equation.

For the minimum mass

*M*, the block is on the verge of sliding up the incline. This means that

*Mg*sin

*θ*—

*T*+

*Mg*cos

*θ**μ*_{s} = 0, where

*T* =

*mg*. We can calculate the minimum

*M* from the previous equation.

__Answer:__ Minimum

*M* =

*m*/(sin

*θ*+cos

*θ**μ*_{s}), Maximum

*M* =

*m*/(sin

*θ*—cos

*θ**μ*_{s})

__Hint and answer for Problem # 6__
This is a challenging problem! It took me a while to figure this one out!

At some angle

*θ*_{1} >

*θ*_{max1} block

*M* will slide down on its own if there is no rope attached. Similarly, at some angle

*θ*_{2} >

*θ*_{max2} block

*m* will slide down on its own if there is no rope attached.

It is known that

*θ*_{max1} = atan(

*μ*_{s1}) and

*θ*_{max2} = atan(

*μ*_{s2}).

If

*θ*_{1} ≤

*θ*_{max1} and

*θ*_{2} ≤

*θ*_{max2} then no sliding occurs. There are three more cases to consider.

*Case 1:*
*θ*_{1} >

*θ*_{max1} and

*θ*_{2} ≤

*θ*_{max2}.

Apply the equilibrium equation to block

*M* in which it is on the brink of sliding down. We have:

*Mg*sin

*θ*_{1}—

*Mg*cos

*θ*_{1}*μ*_{s1}—

*T*_{min1} = 0, where

*T*_{min1} corresponds to the minimum rope tension preventing block

*M* from sliding down. (Note that the system naturally "settles" such that the rope tension

*T* required to stop the block from sliding down is the minimum possible amount). For

*T* <

*T*_{min1} the block slides down. From this equation we get

*T*_{min1} =

*Mg*sin

*θ*_{1}—

*Mg*cos

*θ*_{1}*μ*_{s1}. Call this equation (1).

There is no need to consider block

*M* sliding up since it is an impossibility for

*θ*_{2} ≤

*θ*_{max2} (which means block

*m* cannot slide down which means it cannot pull block

*M* up).

*T*_{min1} must be provided by the block

*m* and must not exceed the maximum rope tension which can be resisted by block

*m* and not be pulled up the incline. This maximum rope tension can be determined from the following equilibrium equation applied to block

*m*:

*mg*sin

*θ*_{2}+

*mg*cos

*θ*_{2}*μ*_{s2}—

*T*_{max2} = 0, from which

*T*_{max2} =

*mg*sin

*θ*_{2}+

*mg*cos

*θ*_{2}*μ*_{s2}. Call this equation (2).

For no sliding

*T*_{min1} ≤

*T*_{max2}. Therefore, from equation (1) and (2) we have the final inequality for this case:

*M*sin

*θ*_{1}—

*M*cos

*θ*_{1}*μ*_{s1} ≤

*m*sin

*θ*_{2}+

*m*cos

*θ*_{2}*μ*_{s2}
*Case 2:*
*θ*_{1} ≤

*θ*_{max1} and

*θ*_{2} >

*θ*_{max2}.

This is the same as case 1, by symmetry. Hence, the final inequality for this case is:

*m*sin

*θ*_{2}—

*m*cos

*θ*_{2}*μ*_{s2} ≤

*M*sin

*θ*_{1}+

*M*cos

*θ*_{1}*μ*_{s1}
*Case 3:*
*θ*_{1} >

*θ*_{max1} and

*θ*_{2} >

*θ*_{max2}.

The blocks will slide together in one direction or the other. To determine the direction we must first calculate the net force pulling down on each block along their respective inclines, as a result of gravity. We do this as follows:

For block

*M*,

*F*_{net1} =

*Mg*sin

*θ*_{1}—

*Mg*cos

*θ*_{1}*μ*_{s1}. And

*F*_{net1} > 0 since

*θ*_{1} >

*θ*_{max1}.

For block

*m*,

*F*_{net2} =

*mg*sin

*θ*_{2}—

*mg*cos

*θ*_{2}*μ*_{s2}. And

*F*_{net2} > 0 since

*θ*_{2} >

*θ*_{max2}.

We now have three sub-cases to consider. The final inequalities for this case will be given within these three sub-cases, as follows.

*Case 3A:*
If

*F*_{net1} =

*F*_{net2} the blocks will not slide.

*Case 3B:*
If

*F*_{net1} >

*F*_{net2}, then

*F*_{net1} ≤

*mg*sin

*θ*_{2}+

*mg*cos

*θ*_{2}*μ*_{s2} for no sliding. Note that

*F*_{net1} is equal to the rope tension, and this rope tension is the minimum required to prevent block

*M* from sliding down the incline.

Hence, for no sliding:

*M*sin

*θ*_{1}—

*M*cos

*θ*_{1}*μ*_{s1} ≤

*m*sin

*θ*_{2}+

*m*cos

*θ*_{2}*μ*_{s2}
*Case 3C:*
If

*F*_{net2} >

*F*_{net1}, then

*F*_{net2} ≤

*Mg*sin

*θ*_{1}+

*Mg*cos

*θ*_{1}*μ*_{s1} for no sliding. Note that

*F*_{net2} is equal to the rope tension, and this rope tension is the minimum required to prevent block

*m* from sliding down the incline.

Hence, for no sliding:

*m*sin

*θ*_{2}—

*m*cos

*θ*_{2}*μ*_{s2} ≤

*M*sin

*θ*_{1}+

*M*cos

*θ*_{1}*μ*_{s1}
We are done!

__Hint and answer for Problem # 7__
Apply the condition of static equilibrium to the block. We have 2

*F*—

*mg*sin

*θ*—

*mg*cos

*θ**μ*_{k} = 0. The term 2

*F* comes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation for

*F*.

__Answer:__ *F* = (1/2)

*mg*(sin

*θ*+

*μ*_{k}cos

*θ*)

__Hint and answer for Problem # 8__
Apply the condition of static equilibrium to the block. We have 2

*F*—

*mg* = 0. The term 2

*F* comes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation for

*F*.

__Answer:__ *F* =

*mg*/2

__Hint and answer for Problem # 9__
Upon close inspection we see that the bottom two pulleys are held up by four segments of rope. The tension in the rope is assumed equal throughout its length (a good assumption for ropes in general since they weigh little). Three of the four rope segments are vertical while the remaining rope segment is at a small angle with the vertical. But for ease of calculation we can treat it as being exactly vertical. Since we are ignoring the mass of the pulleys, the tension in the four rope segments must equal the weight of the mass, in order to satisfy the condition of static equilibrium. Hence, 4

*F*—

*Mg* = 0. We then solve this equation for

*F*.

__Answer:__ *F* =

*Mg*/4

**Bonus Problem**
A conveyor belt carrying aggregate is illustrated in the figure below. A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at an angle

*θ*. To keep the belt in tension a weight of mass

*m* is suspended from the belt, as shown.

Find the point of maximum tension in the belt. You don’t have to calculate it, just find the location and give a reason for it.

You can get the solution for this for $0.99 USD. It's in PDF format. It's available as an instant download through PayPal.

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