Understanding Forces in the Body: A Homework Challenge

Click For Summary

Homework Help Overview

The discussion revolves around a homework problem related to forces in the body, specifically focusing on calculations involving trigonometry and torque. Participants are attempting to understand the application of these concepts in a practical context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the calculation of force components using trigonometric functions and the concept of torque. Questions about the reasoning behind the calculations and the definitions of terms like torque are raised.

Discussion Status

There is an ongoing exchange where some participants are seeking clarification on their calculations and the underlying concepts. Guidance has been offered regarding the need to articulate thought processes and assumptions, but no consensus has been reached on the correctness of the calculations.

Contextual Notes

Participants have noted that the problem lacks specific details, such as the point about which to calculate torque. There is also a mention of the need to show attempts at solutions before receiving help, in line with forum guidelines.

Shanghai Babe
Messages
6
Reaction score
0
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. thanks u :)

Homework Statement





Homework Equations














The Attempt at a Solution

 

Attachments

  • Biophysics- quiz forces in the body.jpg
    Biophysics- quiz forces in the body.jpg
    39 KB · Views: 480
Physics news on Phys.org
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?
 
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.
 
Last edited:
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.
 
@ Simon Bridge spare me the comment! if u don't want to help don't reply!
 
for 1b i got : 50xsin70=46,98 T
 
@ Simon Bridge spare me the comment! if u don't want to help don't 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?

For (c) you have already asked in another thread.
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.
 
Last edited:
i am not here for the pupose that people are doing my homework. i don't 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.
 
Shanghai Babe said:
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?
 
  • #10
Thanks haruspex: the attachment is a little tricky to read isn't it?
Hmmm... my earlier reply doesn't seem to have been recorded - but it was basically what you said with a description of what it means to "show what you are thinking".

@Shanghai Babe :
When you do a problem, you have to think about it ... you'd normally just do it in your head. In order to help you properly, I need to have some idea about this process. From what you've written I can make guesses ... but that will be unlikely to be very pleasing for you and it takes longer.

One way of showing your thinking is to use more plain language descriptions to go with your math - and use symbols more.

It is a useful skill to learn and it can get you more marks when you use it i assignments.
 

Similar threads

Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K