Physics Tension/Equilibrium Problem

[Destrio]

Hey, first post here :)
I got the answer of this tension problem but I don't understand why the angle I used is the angle that works. The physics teacher doesn't know how to explain it either.

Question:
A 4.2m long uniform post is supported by a cable having a tension of 1700N. What is the mass of this post. (diagram included)

My work:
Fcounterclockwise = Fclockwise
1700N * 3.2m = Fp * 2.1m * sin30
Fp = 5.3*10^2 kg

Thanks,
Noah Jordan

EDIT:
until the attachment is up:

the cable is attached 3.2m from the base
and the base of the post is to the right of where the cable is attached to the wall
its at the 30degree angle to the wall
and resting at the 60degree angle (being held up by the cable)
probably not a great description :/

 

[andrevdh]

Welcome to PF Destrio. Hope we can help with some of your physics problems.

I am trying to answer your question without having seen the drawing yet.
These are my assumptions:

It seems that the base of the post is resting on the ground, but the top is not resting against the wall. The tension is perpendicular to the post. You compare the moments of the forces acting on the post about its base. I guess that your problem is with calculating the moment arm, let's call it distance $b$ , of the weight of the post. It is the perpendicular distance from the base of the post to the line of action of the weight of the post. So you need to calculate the base of a right-angled triangle with hypotenuse length $h = l/2 = 2.1\ meter$ .
The angle opposite the base is the angle that the post make with the wall $30^o$. This means that the base will be given by

$$b = h \sin(30^o)$$

 

[kyle8921]

Hi Noah,

First of all, great job on solving the tension problem! It seems like you have a good understanding of the concept. However, I can understand your confusion about the angle used in the calculation.

In this problem, we are trying to find the mass of the post, which is being supported by the cable. The cable is pulling on the post at a 30 degree angle, which means that the force of tension is acting at that angle as well. In order to balance the forces, we need to consider the component of the tension force that is acting in the opposite direction of gravity.

By using the angle of 30 degrees, you were able to find the component of the tension force that is acting in the opposite direction of gravity, which is what we need to balance the forces in this problem. This is why the angle you used in your calculation works.

I hope this explanation helps clarify things for you. Keep up the good work in physics!

Best,