Physics Tension/Equilibrium Problem

  • Thread starter Destrio
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  • #1
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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 :/
 

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Answers and Replies

  • #2
andrevdh
Homework Helper
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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, lets call it distance [itex]b[/itex] , 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 [itex]h = l/2 = 2.1\ meter[/itex] .
The angle opposite the base is the angle that the post make with the wall [itex]30^o[/itex]. This means that the base will be given by

[tex]b = h \sin(30^o)[/tex]
 

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