Statics question about inclined friction

Question

A body of weight (W) is placed on a rough inclined plane makes with the horizontal an angle of measure θ and the force of magnitude (W) acts on it in direction of the line of greatest slope of the plane and the body is about to move up , then μs + tanθ= .........

a) secθ

b) cscθ

c) sinθ

d) cosθ

## I need the solution with detailed steps (why) ##

Answer 1

First of all let's set up the coordinate plane along the slope. Then let's split the weight into two components along that coordinate plane:

The free-body diagram would then look like this:

Where F_F is the force of friction, W is the applied force, F_N is the normal force, Wcos(θ) is the weight in the y direction and Wsin(θ) is the weight component in the x-direction.

Summing the forces in the y-direction, we get the F_N = Wcos(θ)

Summing the forces in the x-direction, we get F_F + Wsin(θ) = W but we also know that the FF = μF_N so the sum of the forces in the y direction becomes:

μF_N + Wsin(θ) = W

Substituting (from the sum of forces in the x-direction) that F_N = Wcos(θ) we get:

μWcos(θ) + Wsin(θ) = W

Dividing through by "W" we get:

μcos(θ) + sin(θ) = 1

Dividing through by cos(θ) we get:

μcos(θ)/cos(θ) + sin(θ)/cos(θ) = 1/cos(θ) or: μ + tan(θ) = sec(θ)

Answer 2

Does "in the direction of the line of the greatest slope" mean up the slope? I assume it does, and I assume you are looking at the time before the block accelerates, and so net forces equal zero and you have static friction with coeff mu_s. (Sorry, not sure how to put Greek letters in here.) Then draw your free body diagram for the body of weight W. You have gravity going straight down, normal force perpendicular to the slope and pointing outward from the surface of the slope. Friction acts against motion, so points down the slope. Another applied force is applied in the direction up the slope and has magnitude equal to the weight of the object. Using geometry, you can show that the angle between the gravity force and the vector pointing opposite to direction of the normal force is equal to theta. Then add all the force components parallel to the slope and set equal to zero. Set sum of all the force components perpendicular to the slope equal to zero. From these two equations, you can solve for your answer. Sorry, it's hard to type all the equations in here, so I described them and hope you can follow. The answer I got for your question is then sec (theta).