Moment of a force about a point

[jonnyboy]

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?

 

[Doc Al]

Use a trig identity to express sinα in terms of cosα (or vice versa). Then you can solve for cosα.

 

[jonnyboy]

can I use sin A = cos(pi/2 - A)?

 

[Doc Al]

Use sin²α + cos²α = 1. (Rearrange it.)