A 12.0 kg object is on a surface that is inclined 30 degrees and the coefficient of friction is 0.65.

Determine the force of static friction if the block remains stationary.

2 answers

Ff = μ m g cos(30º)
Assuming 0.65 is the coefficient of static friction, u.
If you resolve forces acting perpendicular to the ramp, with the direction of the normal force as positive:
Fn - mg*cos30 = 0
Fn - (12)(9.81)*cos30 = 0
Fn = 101.95 N
Then:
Fs = u * Fn
Fs = 0.65 (101.95)
Fs = 66.3 N

Check the opposing force:
mg*sin30
= (12)(9.81)*sin30
= 58.86 N

58.86 N < 66.3 N, which makes sense, because the question states that the block remains stationary ( i.e., there is not enough of an opposing force to overcome the force of static friction).