Forces in body

by Shanghai Babe
Tags: body, forces
 P: 6 hey everybody :) i am new here and i need ur help. I have to solve this problem for home work tomorrow.Topic is forces in the body. I attached it. I actually need help for the whole first question. thx u in advance. Please try to explain it not too difficulty. thx u :) 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution Attached Thumbnails
 HW Helper P: 1,391 Hi! As per the forum guidelines, you need to show us your attempt at the solution before we can help you. What have you tried so far? Where are you getting stuck?
 Homework Sci Advisor HW Helper Thanks P: 12,739 To get the best from this forum, you should show us your best attempt. For that problem (1a) you put axis on the diagram and use your knowledge of trigonometry.
 P: 6 Forces in body i already tried to calculate it. for 1a) i got 17,10N as result. but i am not sure if its right. I calculated : 50xcos70.
 P: 6 @ Simon Bridge spare me the comment!! if u dont want to help dont reply!
 P: 6 for 1b i got : 50xsin70=46,98 T
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P: 12,739
 @ Simon Bridge spare me the comment!! if u dont want to help dont reply!
I cannot help without knowing how you are thinking about the problem.
From the information you supplied, you got a fair comment: but it's a bit terse. I edited so it reads more clearly. Better?

Your numbers don't mean anything - I need to see your reasoning: how are you thinking about the problem.
From what you have written I can make a guess about this - but it is unlikely to be very satisfying for you.

I wonder if you have misunderstood the nature of this place? We can help a student get unstuck but we do not do the work for the student. This creates a problem when someone asks "is this the right answer?" If I just say "yes" or "no" then some students will be cunning and keep guessing until I say "yes" and so get the right answer by making me do the work.

But it appears you can apply the rules of trigonometry OK :)
If ##\theta## is the angle some force ##\vec{F}## to the horizontal, then the horizontal component of that force is, indeed, ##F\cos(\theta)##

You kept going:

 for 1b i got : 50xsin70=46,98 T
You appear to have computed the vertical component of T.
Is that correct?

1b asks for torque
... what is torque?
... what are the units for torque?

But you already know - just consider what the term "translational equilibrium" means.
If in doubt, you can look in your class notes or use google ;)

For (d) it is the same process, but for rotation.
 P: 6 i am not here for the pupose that people are doing my homework. i dont understand what u meant by " what are you thinking about this problem". 50xcos70=17,10N 50 stands for the force (which is given) and 70( is given too) for the angle. relating to 1b) torque is the tendency of a force to rotate the body to which it is applied. torque is always specified with the regard to the axis rotation.relationship betwenn torque and force is : Torque=F(force)x R (radius or distance ) unit of torque is newtonmeters.
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 Quote by Shanghai Babe torque is always specified with the regard to the axis rotation.
It is meaningful (but not always useful) to take the torque about any point. Standing on one end of a non-uniform horizontal beam, each point of the beam is a potential break point, so your torque about each point is worthy of consideration.
The question is a little deficient here. It ought to specify the point to take the torque about, but presumably it means O.
 relationship betwenn torque and force is : Torque=F(force)x R (radius or distance ) unit of torque is newtonmeters.
Quite so (but to clarify, F and R must be as measured at right angles to each other). You gave as answer 50xsin70 T. Not sure what the T represents in that, so I guess you meant N. But as Simon said, that's the vertical component of T, so it's a force, not a torque. What is the radius here?