Question

Box A weighs 80 N and rests on a table (Figure 4-30). A rope that connects boxes A and B drapes over a pulley so that box B hangs above the table, as shown in the figure. The pulley and rope are massless, and the pulley is frictionless. What force does the table exert on box A if box B weighs (a) 35 N; (b) 70 N; (c) 90 N? 64.

Answer 1

Pulley is above the table, the rope goes over the pulley and B hangs above the table, ie does not touch it

B is motionless

So T = 35N, weight of B

So force of rope on A = T = 35 N

Weight of A on table = 80-35 = 45N = Reaction force from table

Answer 2

the net force will tell us ma of the particle

the rope will have a tension which is uniform T

let us consider a) 35N the free body diagram of this weight will have T acting up the rope and 35 N downwards. since this particle is stationary the tension in the rope must be 35N

lets look at the 80N box, what are the forces acting on it ?
there is a tension acting on it up on the box
the weight acts down
the normal reaction acts on the box up
the force acting downards which is just weight should balance the tension and normal reaction
80 = T + N
tension is uniform along the rope we already calculated it as 25 N
80 = 35 + N or N = 45 N

b) 70 N using the same principle of freebody diagrams we will get normal reaction as 9 Newton

c) here since you have a higher weight on the other side, the system is no longer in equilibrium
there is movement
the weight goes down and the rope pulls the 80N weight up
so the box is no longer on the table and hence it cant apply any force on the box
or the normal reaction it exerts will be 0 since there is no normal reaction on the box
even though the question does not ask for it, let me explain
apply the free body diagram on the 90 N
let the tension be T
let it fall down with an acceleration a
90 - T = (90/g) *a net force is mass * accelration
on the other side you have T - 80 = (80/g) * a
here you can solve 31 = 169/g * a or acceleration = 31g/169
and you can solve for T as well