Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin wedge. one edge of the wedge is vertical, and the tip makes an angle of 43⁰. the block that hangs vertically weighs 2 kg, and the block on the incline weighs 2.41 kg.

if the two blocks do not move, what is magnitide of the force of friction on the second second block? fb

Hi, thank you for posting your question here at .

First, you have to draw the system for better understanding. Pls refer to the attached image.

Since the system is not moving, the summation of forces are zero since they are at equilibrium

Let's base on the 2-kg object first to find the tension (T) of the rope.
Forces in the y-direction = 0 = T - W
                                           0 = T - 2 kg (9.81 m/s2)
                                          T = 19.62 N

Let's base on the 2.41 kg load.
Forces in the y-direction = 0 = Fn - Wy
                                           0 = Fn - (2.41kg)(9.81m/s^2)(cos47)
                                          Fn = 16.12 N

Forces in the x-direction = 0 = Ff + T - Wx
                                           0 = Ff + 19.62 - (2.41kg)(9.81m/s^2)(sin47)
                                          Ff = -2.33 N

The magnitude of the frictional force is 2.33 N.