Find the tensions in the cables of a weight

[warfreak131]

Homework Statement

To carry a 100-pound cylindrical weight, two workers lift on the ends of short ropes tied to an eyelet on the top center of the cylinder. One rope makes a 20 degree angle away from the vertical, and the other makes a 30 degree angle.

(a) Find each rope's tension if the resultant force is vertical.

(b) Find the vertical component of each worker's force.

The Attempt at a Solution

Since there is no acceleration, that means that the vertical component of both tensions added together is equal to the weight of the cylinder, and the horizonal components added are equal to 0.

$$F_{T1}Sin70 + F_{T2}Sin60 = 100$$

$$F_{T1}Cos70 + F_{T2}Cos60 = 0$$I'm not sure what to do in these problems, I've never been able to properly answer them.

 

[cepheid]

Your understanding of the physics seems fine. As you've correctly noted, the resultant force being vertical means that the horizontal forces cancel. The problem doesn't actually say that the cylinder doesn't accelerate vertically, but it seems like a good assumption to make, since no vertical acceleration is given. If these assumptions are correct, then you've set up the equations correctly. Now it's just algebra, and fact that the number of equations is equal to the number of unknowns (i.e. two) means that there is a unique solution.

The only thing that seems amiss is that the angles in your equation don't match the angles in your original problem statement.

 

[warfreak131]

Your understanding of the physics seems fine. As you've correctly noted, the resultant force being vertical means that the horizontal forces cancel. The problem doesn't actually say that the cylinder doesn't accelerate vertically, but it seems like a good assumption to make, since no vertical acceleration is given. If these assumptions are correct, then you've set up the equations correctly. Now it's just algebra, and fact that the number of equations is equal to the number of unknowns (i.e. two) means that there is a unique solution.

The only thing that seems amiss is that the angles in your equation don't match the angles in your original problem statement.

arent you supposed to take the angles with respect to the x-axis? in this case the top of the weight?

 

[warfreak131]

I believe I have it:

I had an error in my original post, it should be $$-F_{T1}Cos70+F_{T2}Cos60=0$$, because these forces are opposite.

therefore, $$F_{T1}Cos70=F_{T2}Cos60$$
$$F_{T1}=1.462F_{T2}$$

do the plugging in...
so on and so forth...

$$F_{T2}=44.6$$
therefore
$$F_{T1}=65.268$$

 

[cepheid]

arent you supposed to take the angles with respect to the x-axis? in this case the top of the weight?

It makes no difference whether you measure from the horizontal or the vertical. The angles are complements of each other, so only thing it changes is whether the sine of the angle gives you the y component, or whether the cosine gives you the y component.