While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary. Which of the following statements is always correct regarding the magnitude of the frictional force fs between the bag and slide? (Select all that apply. Here m is the mass of the bag, N is the normal force, μ is the coefficient of friction, and θ is the angle the slide makes with the horizontal.)1. fs = mg cosθ
2. fs = μN
3. fs < μmg
4. fs = mg sinθ
5. fs = N
6. fs = μmg

Fg = m*g (always downward)
N (normal force, always perpendicular to the surface of the slide. going upward)
Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)