A 50 kg block is being held at rest by friction on a ramp angled at 25 degrees above the horizontal. What is the magnitude of the force of friction acting on the block?

Question

A 50 kg block is being held at rest by friction on a ramp angled at 25 degrees above the horizontal. What is the magnitude of the force of friction acting on the block?

Answer 1

Draw a sketch / free-body diagram:

F_g = mg = (50)(9.81) = 490.5 N but this is happening in the negative y-direction so -490.5 N

That vector can be broken down into 2 components F_g-x and F_g-y and using trig ratios:

cos(25°) = adj/hyp = F_g-y/F_g = F_g-y/490.5 therefore F_g-y = -490.5cos(25°) = -444.544 N

sin(25°) = opp/hyp = F_g-x/F_g = F_g-x/490.5 therefore F_g-x = -490.5sin(25°) = -207.294 N

Since there is no motion, ∑F_x = 0 and ∑F_y = 0

∑F_x

F_g-x + F_F = 0

-207.294 + F_F = 0

F_F = 207.294

F_F = 210 N