Vector torque problem: Force applied to a disc

  • #1
3
1

Homework Statement


A force of magnitude 50N is applied at the bottom point Q of a disk of radius 8m that is pinned at P
(leftŸmost point)

See attached picture

(a) Find the angle between the force and the vector from P to Q.
(b) Find the magnitude of the applied torque.



Homework Equations


|T| = |F|sin(Theta)*|r|[/B]

The Attempt at a Solution


So for part (a) I said the angle was 85 degrees but apparently that's wrong? At least according to my colleges online homework program. honestly don't know how to start part (b)
 

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Answers and Replies

  • #2
tnich
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So for part (a) I said the angle was 85 degrees but apparently that's wrong?
Draw the vector from P to Q. Then move it so that its tail is at the same point as the tail of the force vector. Now what do you get for an angle between them?
 
  • #3
You set a usual OXYZ reference system, with origin at point P, and note that the coordinates of point Q are: Q(0,8,-8), and the vector position corresponding to point Q, remains:
r = PQ = <8,-8,0>
Then, note that the components of the force applied at point Q are:
F = <50sin(40º), -50cos(40º), 0>.
Then, to calculate the torque with respect to an axis parallel to the OX axis passing through point P, you have to solve the vector product:
T = r x F.
 
  • #4
3
1
Draw the vector from P to Q. Then move it so that its tail is at the same point as the tail of the force vector. Now what do you get for an angle between them?
95 Degrees?
 
  • #5
tnich
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95 Degrees?
Yes. Now you have everything you need to substitute into your equation for |T|.
 
  • #6
3
1
Yes. Now you have everything you need to substitute into your equation for |T|.
So would I just simply do |T| = 50sin(95)*8sqrt2 or am I missing something?
 
  • #7
tnich
Homework Helper
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336
So would I just simply do |T| = 50sin(95)*8sqrt2 or am I missing something?
I think that would do it.
 

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