The diagram shows a block of mass m = 3.50 kg resting on a plane inclined at an angle of θ = 40° to the horizontal. The coefficient of static friction between the block and the plane is μstatic = 0.30, and the block is stationary but just on the point of sliding down the slope.

Question

Diagram
The diagram shows the four forces acting on the block: an applied force F1 acting perpendicular to the slope (in the –y direction), the block's weight mg, the normal reaction force N and the force of static friction, Ff. In this case, the force of static friction acts up the slope, opposing the tendency of the block to move down the slope.
Find the the minimum magnitude of the applied force F1 that can be exerted if the block is to remain stationary.

Answer 1

M*g = 3.5 * 9.8 = 34.3 N. = Wt. of block.
34.3*sln40 = 22 N. = Force parallel to plane.
34.3*Cos40 = 26.3 N. = Normal force.
Ff = 0.3 * 26.3 = 7.88 N. = Force of static friction.

F1 - Ff = M*a.
F1 - 7.88 = 3.5*0,
F1 = 7.88 N. = Force applied.

Answer 2

Correction: F1 - 7.88 - 22 = 3.5*0.
F1 = 29.88 N.