Question

A 1.56 kg block is held in equilibrium on an incline of angle q = 53.0o by a horizontal force, F, applied in the direction shown in the figure below.

Answer 1

Let:
F be the horizontal force applied,
R be the normal reaction of the plane on the block,
S be the force of friction acting up the plane,
m be the mass of the block,
t be the angle of inclination of the plane to the horizontal,
g be the acceleration due to gravity.

Resolving parallel and perpendicular to the plane:
F cos(t) + S = mg sin(t) ...(1)
F sin(t) + mg cos(t) = R ...(2)

The relationship between S and R is:
S <= uR ...(3)

(a)
Substituting in (3) for S from (1) and for R from (3):
mg sin(t) - F cos(t) <= u(F sin(t) + mg cos(t))
F(cos(t) + u sin(t)) >= mg(sin(t) - u cos(t))
F >= mg[ (sin(t) - u cos(t)) / (cos(t) + u sin(t)) ]
F >= 1.56 * 9.81[ (sin(53) - 0.300 cos(53)) / (cos(53) + 0.300 sin(53)) ]
F > =1.56 * 9.81[ (0.395 - 0.300 (-0.918) / (-0.918) + 0.300 (0.395) ]

F >= 22.91 N.

(b)
From (2):
R = F sin(53) + 1.56 * 9.81cos(53)
R = 22.91 sin(53) + 1.56 * 9.81cos(53)
For the minimum value of F:
R = (.....) N.