Here is one way.Originally Posted byConfuzzled?

Friction.

It tends to go against the direction of motion. In the above problem, it is about static friction.

--If the tendency of the body is to slide down the plane, the static friction will tend to go up the incline. ----------(i)

---If the tendency of the body is to slide up the plane, the static friction will tend to down the incline. ------------(ii)

---Friction depends on the forces normal to the plane. Force due to friction is (normal force) times (coefficient of friction of the plane). -----(iii)

Weight of body is vertical force, it is m*g = 2(9.8) = 19.6 newtons.

External forces parallel to the plane:

Due to weight of body = (19.6)sin(20deg). It is negative because it is going down.

Due to Y = Y*sin(45-20 deg) = Y*sin(25deg). It is positve because it is going up.

So, resultant of forces parallel to plane = [Y*sin(25deg) -19.6sin(20deg)] --------***

External forces normal or perpendicular to the plane:

Due to weight of body = (19.6)cos(20deg).

Due to Y = Y*cos(25deg).

So, resultant of forces normal to the plane = [Y*cos(25deg) +19.6cos(20deg)] --------***

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a) Now if the body tends to slip up the plane.

The friction force tends to go down the plane, or it is negative.

Because the body is still at rest, the sum of all forces parallel to the plane is zero.

So,

[Y*sin(25deg) -19.6sin(20deg)] -(0.2)[Y*cos(25deg) +19.6cos(20deg)] = 0

Y[sin(25deg) -0.2cos(25deg)] -19.6[sin(20deg) +0.2cos(20deg)] = 0

Y[0.241356704] -19.6[0.529958667] = 0

Y = 19.6(0.529958667)/(0.241356704) = 43 newtons ---------answer.

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b) If the body tends to slip down the plane.

The friction force tends to go up the plane, or it is positive.

Because the body is still at rest, the sum of all forces parallel to the plane is zero.

So,

[Y*sin(25deg) -19.6sin(20deg)] +(0.2)[Y*cos(25deg) +19.6cos(20deg)] = 0

Y[sin(25deg) +0.2cos(25deg)] -19.6[sin(20deg) -0.2cos(20deg)] = 0

Y[0.603879819] -19.6[0.154081619] = 0

Y = 19.6(0.154081619)/(0.603879819) = 5 newtons ---------answer.