Tension and Pulleys! (Statics of Particles)

**jegues** · January 27th 2010, 04:39 PM

1. The problem statement, all variables and given/known data
See figure.

2. Relevant equations
Sum of Fy = 0
Sum of Fx = 0

3. The attempt at a solution

I can't see how to solve this. The only thing I can think of is the vertical forces of both sides of the pulley are going to equal the gravitational force of the mass in and that the two x components on each side of the rope will cancel.

I don't have the magnitude on the right hand side of the pulley and on the left hand side I don't have the magnitude or the angle. How do we solve this?

Thanks in advance. (If I posted this in the wrong forum please let me know)

**Grandad** · January 28th 2010, 04:34 AM

Hello jegues

Welcome to Math Help Forum!

Originally Posted by jegues

1. The problem statement, all variables and given/known data
See figure.

2. Relevant equations
Sum of Fy = 0
Sum of Fx = 0

3. The attempt at a solution

I can't see how to solve this. The only thing I can think of is the vertical forces of both sides of the pulley are going to equal the gravitational force of the mass in and that the two x components on each side of the rope will cancel.

I don't have the magnitude on the right hand side of the pulley and on the left hand side I don't have the magnitude or the angle. How do we solve this?

Thanks in advance. (If I posted this in the wrong forum please let me know)

If the pulleys are smooth and frictionless (and we must assume they are) then the tension in the string is the same throughout its length. Since the tension at the free end is $T$ , then the tension throughout will be $T$ . So we have a force $2T$ acting at $27^o$ to the vertical on the RHS.

So use the equations you have written down in (2), and solve for T and $\alpha$ .

Here's the first one:

$2T\sin27^o+T\cos\alpha = 0$

Answers:

Spoiler:

Grandad

**jegues** · January 28th 2010, 08:17 AM

So,

$ 2T\sin27^o+T\cos\alpha = 0 $

and

$ 2T\cos27^o+T\sin\alpha = 0 $

How do you go about solving this?

Is it simply,

$ T(2\sin27^o+\cos\alpha) = 0 $

T cannot be zero so we solve for the part in brackets?

Thanks again!

**Grandad** · January 28th 2010, 09:43 AM

Hello jegues

Originally Posted by jegues

So,

$ 2T\sin27^o+T\cos\alpha = 0 $

and

$ 2T\cos27^o+T\sin\alpha = 0 $

How do you go about solving this?

Is it simply,

$ T(2\sin27^o+\cos\alpha) = 0 $

T cannot be zero so we solve for the part in brackets?

Thanks again!

Yes, but (apologies!) the equation should read

$2T\sin27^o - T\cos \alpha=0$ (Of course!)

So you get:

$\cos\alpha = 2\sin27^o$

$\Rightarrow \alpha = 24.8^o$

But your second equation isn't right. You've forgotten the weight force, which is $-200\times 9.8$ N. Add this into your equation, substitute the value you have found for $\alpha$ , and solve for $T$ .

Grandad

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