The mass of the block in the drawing is 10 kg. The coefficient of static friction between the block and the vertical wall is 0. 50. What minimum force f is required to start the block moving up the wall?.

It depends on the direction of the applied force. Consult the attached free body diagram.

If the block is held in place by a force acting perpendicular to the wall, then the block will never move. So the applied force must be acting at some angle θ between 0° and 90° (with respect to the positive horizontal axis).

By Newton's second law, we have

• net perpendicular force

∑ F (⟂) = F cos(θ) - N = 0

• net parallel force

∑ F (//) = F sin(θ) - f - mg = ma

However, we're interested in the minimum force required to get the block moving. This minimum force would exactly counter the frictional force keeping the block in place, so that

F sin(θ) - f - mg = 0

The magnitude of friction is 0.5 N (that is, half the normal force's magnitude, not 0.5 Newtons). So

F sin(θ) - 0.5 N - mg = 0   ⇒   F sin(θ) - 0.5 F cos(θ) = mg

⇒   F (sin(θ) - 0.5 cos(θ)) = mg

⇒   F = mg/(sin(θ) - 0.5 cos(θ))

(where m = 10 kg and g = 9.8 m/s²)