The figure shows a block P on top of another block Q. The two blocks are connected by a string which passes over a fixed smooth pulley. The mass of P is 0.20 kg and the mass of Q is 0.50 kg. The coefficient of static friction between blocks P and Q is 0.5 and between Q and the floor is 0.2. What is the maximum horizontal force F which can be applied before relative motion occurs between block Q and the floor. (3.335N)
The foecr balance three forces:
a. max friction between Q and floor F1=(0.5+0.2)*0.2*9.8
b. max friction between Q and P F2=0.2*0.5*9.8
c. max string tension F3=F2
finally F=F1+F2+F3=0.34*9.8=3.3 N
To find the maximum horizontal force F that can be applied before relative motion occurs between block Q and the floor, we need to consider the forces acting on block Q.
Let's break down the forces acting on block Q:
1. The gravitational force (weight) acting vertically downwards with a magnitude of mQ * g, where mQ is the mass of block Q and g is the acceleration due to gravity.
2. The normal force from the floor acting vertically upwards with a magnitude equal to mQ * g, as both blocks are at rest.
3. The frictional force between block Q and the floor acting horizontally in the opposite direction of the applied force. The maximum static frictional force is given by the coefficient of static friction (μ) multiplied by the normal force.
Given that the coefficient of static friction between block Q and the floor is 0.2, the maximum static frictional force is 0.2 * (mQ * g).
Now, for block Q to remain stationary, the applied force (F) must be less than or equal to the maximum static frictional force. Therefore, we have:
F ≤ μ * (mQ * g)
Substituting the values we have:
F ≤ 0.2 * (0.50 kg * 9.8 m/s^2)
Calculating this expression:
F ≤ 0.2 * 4.9 N
F ≤ 0.98 N
Therefore, the maximum horizontal force F that can be applied before relative motion occurs between block Q and the floor is 0.98 N.