Question

How do you solve this if mass isn't known??? (Forces question)

This is a long one on forces. The question is asking,

a) At `t = 0`, what is the total force `F` acting on the suitcase? Include a force diagram.

b) How long does the suitcase take to reach the speed of the conveyor belt, i.e. at what time `t` does `v(t)` = `v_b`? Express your answer in terms of `g`.

c) After the suitcase reaches the speed of the conveyor belt, what is the force of friction that acts on it?

Thing here is, luggage has mass M, static friction between conveyor and luggage and it's kinetic is given: `u_k < u_s`. Conveyor moves with speed `v_b` and time `t=0` and you place the luggage on the belt with speed `v=0`. Now I'm used to seeing a mass value of some sort and other values. I assume the total Force for part a at least is `= 0` as `T=0` and `v=0`.

Answer 1

Let the movement of conveyor belt be along the `x`-axis and upwards direction is of `z`-axis.

(a) At `t=0`, as the suitcase is placed on the conveyor belt three forces are acting on the suitcase.

1. Weight acting downwards `=-Mghatz`.
2. Normal reaction acting upwards `=Mghatz`
3. Force of friction `=u_kMghati`
   We know that force of friction always acts in a direction to oppose the motion. In this case, force of friction is responsible for accelerating the suitcase so that it attains the speed of the conveyor belt. As the movement of the belt is along `+ve` `x`-axis, therefore direction of force of friction is of `hati`
   (Please draw a Force diagram to complete the answer)

(b) Assuming acceleration to be uniform we get from Newton's Second Law of motion

`a=(u_kMg)/M=u_kg`

Using kinematic expression

`v=u+at`

we get

`v_b=0+u_kg t`
`=>t=v_b/(u_kg)`

(c) Once the suitcase reaches the constant speed of the conveyor, we see that there is no acceleration of the suitcase. Therefore, as is evident from the Newton's First law of motion, force of friction has to be zero.