Question

A 1000-kg load is suspended from the end of a horizontal boom. The boom is supported by a cable that makes an angle of 35 degrees with the boom.
a) What is the weight of the load?
b) What is the tension in the cable?
c) What is horizontal force on the boom?
d) What is the vertical equilibrium component of the tension in the cable?

Diagram here: http://highered.mcgraw-hill.com/sites/dl/free/0070735824/582661/mhr_calculus_sample_ch06.pdf
pg. 53, question #9

I figured out question a and got help with b, however I'm stil unsure of how to figure out question c

Answer 1

Isn't this a physics question?

Answer 2

I guess it is technically physics, but we're doing vectors in my Calculus course right now.

Answer 3

I'm sorry, I don't remember highschool physics very well. All those tension formulas escape me. Let me find someone who can help you. Meanwhile, you can go through this,
http://www.physicsforums.com/showthread.php?t=270928

Answer 4

Thank you!

Answer 5

The tension in the cable contains a force component that must balance out the downward force of the 1000kg weight at the end of the boom (the system is in static equilibrium). The vertical component of the tension vector is T*sin(35) = 1000kg*g where g is acceleration due to gravity (9.81 m/s^2) and T is the tension you need to find. Thus, the magnitude of tension in the cable is 9810/sin(35) Newtons.

Answer 6

Btw, I did not look at the diagram. I am assuming that the boom is pinned on the end opposite of the 1000 kg load. Otherwise, the answer might be different.

Answer 7

Yes, it its pinned on the side opposite the load. That makes sense, thank you!