Equilibrium of a particle

Question

a particle whose weight W rests on a smooth inclined plane under the action of a horizontal force P. The particle can also be kept in equilibrium on the plane by applying a force Q along the plane upwards. Show that 1/Q^2 = 1/P^2 + 1/W^2

Answer 1

For the gravitational force, that starts as W, acting vertically (note the distinction from mass; this is already converted into a force!). That decomposes into W sin θ acting down the plane, and W cos θ acting into the plane.

Next, the horizontally acting force P, decomposes into P cos θ acting up the plane, and P sin θ acting into the plane.

Since P could balance W, then W sin θ = P cos θ.

Lastly, Q is totally along the plane already, so if it balances W, Q = W sin θ. Equating these equalities, and ^2ing:

W^2 sin^2 θ = P^2 cos^2 θ = Q^2; then

W^2/P^2 sin^2 θ = cos^2 θ = Q^2/P^2

likewise sin^2 θ = P^2/W^2 θ = Q^2/W^2

sin^2 θ + cos^2 θ = Q^2/P^2 + Q^2/W^2 = 1 by trig definition; divide through by Q^2:

1/P^2 + 1/W^2 = 1/Q^2

Answer 2

The slope has an elevation angle of Θ. The tangential component of the weight is Wsin(Θ). This is given as Q.

If a force P is applied horizontally, it resolves to:

Pt = P/cos(Θ), and Pn = P/sin(Θ). (we are concerned only with Pt

We then have:

Q = Wsin(Θ)

P = Q/cos(Θ)

Square each side:

Q² = W² * sin²(Θ)

P² = Q² / cos²(Θ)

Solve each one for sin² or cos² , and set the sum = to 1

Q² / W² + Q² / P² = 1

Divide both sides by Q² :

1 / W² + 1 / P² = 1/ Q²