# Pulley System and Applied Force

#### altitude

##### Full Member
My question is about TBR in-chapter question 2.16a on a 2-pulley system and the applied force to support a mass. The picture below is similar to the one for this question; but, in TBR, the weight of the mass is m (rather than 100 kg), and the applied force to the diagonal downward part of the rope is applied force (Fa) (rather than 50 kg):

View attachment Picture 8.png

The question asks how much force must be applied (Fa) to support the mass (m) (assuming the roes and pulleys are massless and frictionless). The applied force is to the diagonal downward part of the rope (as in the picture above).

The options:
A. 2mg
B. mg
C. mg/2
D. mg/3

Their explanation:
-Isolating the lower pulley, which is in equilibrium (i.e. zero acceleration): downward force of mg and upward force of 2T (tension T in 2 vertical parts of the rope); so, 2T = mg; T = mg/2
-If Fa just supports the mass then the applied force is just the tension; so, since T = mg/2, the applied force to the diagonal downward part of the rope is mg/2.

What I don't understand:
-Wouldn't using the tension in the vertical part of the rope give the y-components of tension for the diagonal downward part of the rope?
-Since the question is asking for the applied force/tension in the diagonal part of the rope, shouldn't the answer be:
Applied force is diagonal part of rope = T / sin(theta) (or T / cos(theta), depending on which angle is theta); let's use Fa = T / sin(theta)
Since, T = mg/2; applied force = mg / 2sin(theta)
Doesn't TBR's answer assume that the angle theta is 90 degrees?

Another way to look at it (see picture):
Since the pulley system is in equilibrium, the downward force on the right side of the upper pulley cancels out with the upward force on the left side of the lower pulley. So, what cancels out the upward force on the right side of the lower pulley, is the y-components of the downward force on the left side of the upper pulley. So, wouldn't the applied force on the diagonal downward part of the rope need to take this y-component aspect into account?

Last edited:

#### Jepstein30

##### Full Member
My question is about TBR in-chapter question 2.16a on a 2-pulley system and the applied force to support a mass. The picture below is similar to the one for this question; but, in TBR, the weight of the mass is m (rather than 100 kg), and the applied force to the diagonal downward part of the rope is applied force (Fa) (rather than 50 kg):

View attachment 20431

The question asks how much force must be applied (Fa) to support the mass (m) (assuming the roes and pulleys are massless and frictionless). The applied force is to the diagonal downward part of the rope (as in the picture above).

The options:
A. 2mg
B. mg
C. mg/2
D. mg/3

Their explanation:
-Isolating the lower pulley, which is in equilibrium (i.e. zero acceleration): downward force of mg and upward force of 2T (tension T in 2 vertical parts of the rope); so, 2T = mg; T = mg/2
-If Fa just supports the mass then the applied force is just the tension; so, since T = mg/2, the applied force to the diagonal downward part of the rope is mg/2.

What I don't understand:
-Wouldn't using the tension in the vertical part of the rope give the y-components of tension for the diagonal downward part of the rope?
-Since the question is asking for the applied force/tension in the diagonal part of the rope, shouldn't the answer be:
Applied force is diagonal part of rope = T / sin(theta) (or T / cos(theta), depending on which angle is theta); let's use Fa = T / sin(theta)
Since, T = mg/2; applied force = mg / 2sin(theta)
Doesn't TBR's answer assume that the angle theta is 90 degrees?

Don't think in vertical or horizontal components for pulley systems. The tension is equally distributed regardless what angle you pull on.

#### altitude

##### Full Member
Don't think in vertical or horizontal components for pulley systems. The tension is equally distributed regardless what angle you pull on.

In TBR, on the page before this question they have an example with a 2-pulley system with a mass hanging from a lower pulley, which is supported by two upward diagonal ropes (one going upwards to the left and one upwards to the right). And the formula in TBR for this system is 2Tsin(theta) = W, where they use sin to designate the force of tension in the y-direction to satisfy "the sum of forces in the y-direction Fy = 0" so, "Fy = 2Tsin(theta) - W = 0".

#### Jepstein30

##### Full Member
In TBR, on the page before this question they have an example with a 2-pulley system with a mass hanging from a lower pulley, which is supported by two upward diagonal ropes (one going upwards to the left and one upwards to the right). And the formula in TBR for this system is 2Tsin(theta) = W, where they use sin to designate the force of tension in the y-direction to satisfy "the sum of forces in the y-direction Fy = 0" so, "Fy = 2Tsin(theta) - W = 0".

Right, but the theta there is the angle between the mass and the rope itself, right?

Post that picture if you can.

#### Jepstein30

##### Full Member
I'll wait until you post the other picture but draw a force diagram.

In your first example, you have the Tension pulling straight up TWICE (two ropes) and then weight (mg) going down. Therefore, in equilibrium 2T=mg or T = mg/2.

In your second example, you have two ropes pulling one mass.. one up to the left, one up to the right.

So the force diagram here would have one Tension up to the left and one tension up to the right. Now, we have an x and y component.. before we ONLY had a y component.

Draw a force diagram, ignore "how" you have to pull on the rope to add the tension though.

#### sciencebooks

##### Full Member
10+ Year Member
Right, but the theta there is the angle between the mass and the rope itself, right?

Post that picture if you can.

Hey Jepstein, quick question. In multiple pulley systems, only the pulleys NOT attached to the ceiling contribute to lowering the force by one half, correct? (Lol, sorry, does that even make sense?) Or do you just think of it by counting the number of ropes with tension and then divide force by that number?

#### Jepstein30

##### Full Member
Hey Jepstein, quick question. In multiple pulley systems, only the pulleys NOT attached to the ceiling contribute to lowering the force by one half, correct? (Lol, sorry, does that even make sense?) Or do you just think of it by counting the number of ropes with tension and then divide force by that number?

I draw a force diagram, putting in as many tensions "upward" as there are ropes coming out of the mass..

I'm not very confident with pulley systems being simple ones though so not sure if that's really the way to go. I don't think what you said is right though because then the pulley system above wouldn't reduce the force at all.

I'm pretty sure its just ropes going up.

#### sciencebooks

##### Full Member
10+ Year Member
I draw a force diagram, putting in as many tensions "upward" as there are ropes coming out of the mass..

I'm not very confident with pulley systems being simple ones though so not sure if that's really the way to go. I don't think what you said is right though because then the pulley system above wouldn't reduce the force at all.

I'm pretty sure its just ropes going up.

Yeah, that makes sense. I think it's the way TPR explains sometimes, like *they* just see a correlation and use that in the answer instead of the conceptual reason why that works (if that makes sense). Like a passage said something like "There are 8 pulleys, but 4 are attached to the ceiling so F is reduced only by 1/(2x4)." But cool, thanks for the clarification.

#### sciencebooks

##### Full Member
10+ Year Member
The picture for my question (see post #1):
View attachment 20434

The picture for other example (see post #3):
View attachment 20435

(Disclaimer: these pictures belong to TBR and are posted to better explain my question)

I believe this is because in the second photo, the ropes supporting that pulley are actually positioned at an angel where as in the first photo, they're not.

#### Jepstein30

##### Full Member
The picture for my question (see post #1):
View attachment 20434

The picture for other example (see post #3):
View attachment 20435

(Disclaimer: these pictures belong to TBR and are posted to better explain my question)

Right, so draw force diagrams for both of these.

In the first picture, you have T straight up and mg straight down. This tension only has a vertical component.

In the second picture, you have T at an angle theta to the left and T at an angle theta to the right. These tensions have both a horizontal and vertical component.

#### altitude

##### Full Member
I believe this is because in the second photo, the ropes supporting that pulley are actually positioned at an angel where as in the first photo, they're not.

So, as a rule, only the ropes that are supporting the pulley with the mass need to be broken into vector components? Whereas, the direction of ropes not in direct contact with the pulley supporting the mass is insignificant?

Could someone else confirm this?

#### altitude

##### Full Member
Right, so draw force diagrams for both of these.

In the first picture, you have T straight up and mg straight down. This tension only has a vertical component.

In the second picture, you have T at an angle theta to the left and T at an angle theta to the right. These tensions have both a horizontal and vertical component.

So, are you saying that in the first picture, because there is no x-component, the part of the rope where the force is applied only as a y-component (and its x-component is zero), so the applied force in the diagonal direction is equal to the force in the y-direction?

#### Jepstein30

##### Full Member
So, are you saying that in the first picture, because there is no x-component, the part of the rope where the force is applied only as a y-component (and its x-component is zero), so the applied force in the diagonal direction is equal to the force in the y-direction?

Yes because the tension anywhere in the rope is equal.

So regardless at which angle you pull on it, if you apply force X, you will get force X everywhere else in the rope. Since in this case, you only have a vertical component.. the vertical component is of magnitude X.

If you had a horizontal component, it'd be broken down appropriately.

#### sciencebooks

##### Full Member
10+ Year Member
So, as a rule, only the ropes that are supporting the pulley with the mass need to be broken into vector components? Whereas, the direction of ropes not in direct contact with the pulley supporting the mass is insignificant?

Could someone else confirm this?

I've never seen it so explicitly stated, but I believe that's the gist of it. No matter what angle the person pulls at, they're going to need to give it a force = the force of the tension.

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