Vector torque problem: Force applied to a disc

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SUMMARY

The discussion focuses on calculating the torque applied to a disk when a force of 50N is exerted at point Q, located at the bottom of the disk with a radius of 8m. The angle between the force vector and the vector from point P to Q was initially miscalculated as 85 degrees but was corrected to 95 degrees. The torque magnitude is determined using the equation |T| = |F|sin(Theta)*|r|, where the correct values are substituted to find the torque.

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jwbensin
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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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jwbensin said:
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?
 
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.
 
tnich said:
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?
 
jwbensin said:
95 Degrees?
Yes. Now you have everything you need to substitute into your equation for |T|.
 
tnich said:
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?
 
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jwbensin said:
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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