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# Help finding an unknown force on an inclined plane? watch

1. Part (a) was to find tension without P, which is 694N. I've drawn a diagram so far for (b)

Can someone tell me what do I next? I was going to resolve parallel to the slope: P-110gsin(40.1)-T+T=110(0) but don't think thats right?
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2. (Original post by ℓove)

Part (a) was to find tension without P, which is 694N. I've drawn a diagram so far for (b)

Can someone tell me what do I next? I was going to resolve parallel to the slope: P-110gsin(40.1)-T+T=110(0) but don't think thats right?

Resolving parallel to the slope is a good decision, however:

1) Considering the forces on the mass, you will have T acting up the slope (and that will be 550N when rope is about the snap), but there is no minus T force acting on the mass.

2) The force P is horizontal, not parallel to the slope. It looks as if you drew it correctly the first time, but then erased it.
3. (Original post by ghostwalker)
Resolving parallel to the slope is a good decision, however:

1) Considering the forces on the mass, you will have T acting up the slope (and that will be 550N when rope is about the snap), but there is no minus T force acting on the mass.

2) The force P is horizontal, not parallel to the slope. It looks as if you drew it correctly the first time, but then erased it.
Thanks I've fixed my diagram. Can you tell me what's missing from this equation: P-110gsin(40.1)+T=110(a) I got 144N but the answer is 189N?
4. (Original post by ℓove)
Thanks I've fixed my diagram. Can you tell me what's missing from this equation: P-110gsin(40.1)+T=110(a) I got 144N but the answer is 189N?
P is at an angle to the slope. The comonent of P parallel to the slope is Pcos(40.1), and that's what you need in your equation.
5. (Original post by ghostwalker)
P is at an angle to the slope. The comonent of P parallel to the slope is Pcos(40.1), and that's what you need in your equation.
I divided the answer by cos40.1 and got 189N
Thanks!!!!

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