A 60.0 N force is applied at an angle of theta = 35.0degree to a 8.00 kg block p

Question

A 60.0 N force is applied at an angle of theta = 35.0degree to a 8.00 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. Calculate the coefficient of kinetic friction between the block and the wall. Draw a FBD of the block. Find the normal force as a function of F (an equation - not a number), the unknown applied force, using Newton's Second Law in the horizontal direction. Find the kinetic friction force as a function of F (an equation - not a number). Use Newton's Second Law in the vertical direction and algebra to calculate mu_t.

Explanation / Answer

a) gravity acts downward, normal force acts towards left and kinetic friction acts upward

b) Let N is the Normal force on the block.

in horizontal direction, net force acting block, Fnetx = 0

F*cos(theta) - N = 0

==> N = F*cos(theta)

c) now Apply, Fnety = 0

F*sin(theta) + kinetic Friction - m*g = 0

kineric friction = m*g - F*sin(theta)

d)

kineric friction = m*g - F*sin(theta)

N*mue_k = m*g - F*sin(theta)

F*cos(theta)*mue_k = m*g - F*sin(theta)

mue_k = m*g/(F*cos(theta)) - tan(theta)

= 8*9.8/(60*cos(35)) - tan(35)

= 0.895

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