I suppose you mean to say that the wall is frictionless in the first scenario? Also, I assume the block is to held in place. Construct a free body diagram for the block. There are 3 (in part A) or 4 (in part B) forces acting on it.
• its weight w, pulling it downward
• the normal force (magnitude n), pushing outward from the wall to the left
• the push as described, with magnitude p
• static friction (mag. f ), opposing the upward net force and thus pointing downward.
The static friction force has a magnitude proportional to that of the normal force. If the coefficient of static friction is µ, then
f = µ n
so if the wall is frictionless with µ = 0, then f = 0 and does not need to be considered.
(A) If µ = 0, then by Newton's second law we have
• net vertical force:
∑ F = p sin(50°) - w = 0
and we don't need to consider the net horizontal force to determine p from here. We get
p = w / sin(50°) = (4 kg) (9.8 m/s²) / sin(50°) ≈ 51 N
(B) If µ = 0.25, then Newton's second law gives
• net vertical force:
∑ F = p sin(50°) - w - f = 0
p sin(50°) - f = (4 kg) (9.8 m/s²)
p sin(50°) - f = 39.2 N
• net horizontal force:
∑ F = p cos(50°) - n = 0
p cos(50°) - f /0.25 = 0
[since f = 0.25 n]
p cos(50°) - 4f = 0
Multiply the first equation by -4, then add it to the second equation to eliminate f and solve for p :
-4(p sin(50°) - f ) + (p cos(50°) - 4f ) = -4 (39.2 N) + 0
p (cos(50°) - 4 sin(50°)) = -156.8 N
p = (156.8 N) / (4 sin(50°) - cos(50°)) ≈ 65 N
