Posted by **Stephanie** on Friday, April 22, 2011 at 9:35am.

A 100-kg box is suspended from two ropes; the left one makes an angle of 20degrees with the vertical, and the other makes an angle of 40degrees. What is the tension in each rope. I think I've got the basics:

I have Fx=T1(x) + T2(x)= 0 to calc the first tension, and Fy= T1(y) + T2(y) + w = 0 for the second tension.

To get T1(x)and T2(x) I took cosine of angle 20 and angle 40 to get -0.940 and 0.766. For the T1(y) and T2(y) I took sine of each angle to get 0.342 and 0.643. I converted mass of 100kg to -980N. Do I have that part right? And can you help me the rest of the way. I have spent two hours on this already. Thank you so much.

After this question, the only other one I'm stuck on is a ladder question. You have similar ones on this site, but not the same and I cant figure out what equation I should be using: A 10-kg ladder 2.5m long rests against a frictionless wall with its base on the floor 80cm from the wall. how much force does the top of the ladder exert on the wall?

- Physics -
**drwls**, Friday, April 22, 2011 at 10:57am
Let's discuss the box and rope question. You should post the other one separately.

Since the angles that you were given are measured from vertical, you should use sine for the horizontal (x) components and cosine for the vertical (y) components. You have them mixed up.

You should write you equations in terms of T1 and T2, the two unknown tension forces you want to calculate.

When you substitute sines and cosines, you end up with

T1*sin 20 - T2*sin40 = 0

and

T1*cos 20 + T2*cos 40 = W = M g

0.3420 T1 - 0.6428 T2 = 0

0.9397 T1 + 0.7660 T2 = 100 * g

Don't forget to multiply the mass by g = 9.8 m/s^2 to get the tensions in newtons.

The rest is just algebra. Try substituting

T1 = 1.8794 T2

into the last equation and solving for T2

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