Moment of a force about a point

  • Thread starter jonnyboy
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  • #1
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Homework Statement


This one's easy. A 2.9 lb force (P) is applied to a lever. Determine the value of alpha knowing that the moment of P about A is counterclockwise (+) and has a magnitude of 17 lb-in.


Homework Equations



Moment about a = (Force of x-component)(distance from a to Force point) + (F_y-comp)(dist. to foce)

M_a = (F_x)(d_y) + (F_y)(d_x)

Answer: alpha = 49.9 degrees or 59.4 degrees

The Attempt at a Solution



17 = (2.9sin(alpha))(4.8) + (2.9cos(alpha))(3.4)

how do I solve for alpha?
 

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Answers and Replies

  • #2
Doc Al
Mentor
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Use a trig identity to express sinα in terms of cosα (or vice versa). Then you can solve for cosα.
 
  • #3
18
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can I use sin A = cos(pi/2 - A)?
 
  • #4
Doc Al
Mentor
45,010
1,286
Use sin2α + cos2α = 1. (Rearrange it.)
 

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