How Do You Calculate the X and Y Components of Forces in a Pully System?

[Rob K]

Hi guys,

Got a general question here about sum of forces that keeps coming up for me, and so I want to try and nail it this time.

If you have a pully system, and there is a weight attached to a rope at one end of say 10 kg and the other end is is attached to a force of 1000 N that is pulled at an angle of let's say 40˚ below the horizontal, I am looking to find the x and y components of the forces at the origin. Friction is negligable

So this is what I think it should be:

Using $\Sigma$F = ma

Acceleration due to gravity (g) = 9.81m/s²,
Tension in the rope = T

+$\uparrow$ $\Sigma$F_y
-10kg * 9.81m/s² + T = -10a + 1000cos40N

Where a is yet to be derived

+$\rightarrow$$\Sigma$F_x
? = 1000sin40.

I'm pretty sure I have got these wildly wrong, but my major focus on this question is what needs to be there to balance out the x direction. I am assuming that it is some force acting on the pivot, but how is that expressed. And what other forces or data am I missing from this to make it a viable question.

Kind regards

Rob K

 

[zhermes]

I'm not entirely sure about the geometry of your problem, but your basic approach is correct.

Separate the forces into orthogonal directions, and sum to find (or constrain) the net forces. Remember you need to consider each body independently.

In your equation for the y-direction: there are only two forces on the hanging object---gravity and tension. The effect of the pulley is to transfer the angled-force, to a directly upwards force on the weight----thus it doesn't really matter what the angle is. Additionally, the weight feels no horizontal forces.

 

[Rob K]

Yes, that makes a lot of sense.

The thing is if all the forces in either the y or the x-axis equal each other, here, what would be the equaling force on the x axis? This is where I get lost, what balances out 1000 sin 40 to make the sum of forces in the x-axis balance. Ok if you were in an exam what would you call it to get a mark?

 

[PhanthomJay]

Hi guys,

Got a general question here about sum of forces that keeps coming up for me, and so I want to try and nail it this time.

If you have a pully system, and there is a weight attached to a rope at one end of say 10 kg and the other end is is attached to a force of 1000 N that is pulled at an angle of let's say 40˚ below the horizontal, I am looking to find the x and y components of the forces at the origin. Friction is negligable

So this is what I think it should be:

Using $\Sigma$F = ma

Acceleration due to gravity (g) = 9.81m/s²,
Tension in the rope = T

+$\uparrow$ $\Sigma$F_y
-10kg * 9.81m/s² + T = -10a + 1000cos40N

Where a is yet to be derived

This equation is not correct. The vertical components of the tension forces on the pulley must be balanced by the vertical component of force at the pivot point (from a bolt or cord) on the pulley. There is no net vertical force on the pulley, because it is not moving vertically.

+$\rightarrow$$\Sigma$F_x
? = 1000sin40.

your sines and cosines are reversed...x comp is Tcos40

I'm pretty sure I have got these wildly wrong, but my major focus on this question is what needs to be there to balance out the x direction. I am assuming that it is some force acting on the pivot, but how is that expressed. And what other forces or data am I missing from this to make it a viable question.

The force of the pivot is expressed as the reaction force to the pulley loads. Draw free body diagrams!

 

[alphy]

Hi Rob,

It looks like you are on the right track with your approach to solving this problem. In order to find the x and y components of the forces at the origin, you will need to use trigonometry to break down the force vector into its x and y components. This will give you the following equations:

\begin{align*}
\Sigma F_x &= T \sin(40^\circ) \\
\Sigma F_y &= T \cos(40^\circ) - 10kg * 9.81m/s^2
\end{align*}

From here, you can use the equation \Sigma F = ma to solve for the acceleration, which will give you the necessary information to find the x and y components at the origin.

As for the forces at the pivot, you are correct in assuming that there will be a force acting on the pivot to balance out the x direction. This force is known as the normal force and is equal in magnitude to the x component of the tension force. So your final equations should look something like this:

\begin{align*}
\Sigma F_x &= T \sin(40^\circ) = ma \\
\Sigma F_y &= T \cos(40^\circ) - 10kg * 9.81m/s^2 = 0 \\
N &= T \sin(40^\circ)
\end{align*}

I hope this helps you in solving your problem. Good luck!