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).
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).
2 answers