Newton's Second Law and Friction

[am13]

Homework Statement

Block A weighs 1.14 and block B weighs 3.66. Block A rests on top of block B and has a cable connecting it to the wall on its right. The coefficient of kinetic friction between all surfaces is 0.300. Find the magnitude of the horizontal force necessary to drag block B to the left at constant speed if A is held at rest. (See Figure B)

Homework Equations

Fnet = ma
F_f=μN
Fnet = 0

The Attempt at a Solution

I think that I am just getting the forces on the free body diagram wrong. I drew FBDs for block B and block A and got that Fnet = 0, and so the horzontal force plus force friction by block A minus force friction by the ground equals 0. And I got F = Ff_G-Ff_A. What have I done wrong?

 

[rl.bhat]

Hi am13, welcome to PF.
The frictional force acting on A is towards right.
Reaction to this force on B acts towards left.
Applied force on B is towards left.
Frictional force on B acts towards...?
What is the condition for the constant velocity of B?

 

[tomcue]

As a scientist, it is important to ensure that all the forces acting on the system are accurately represented in the free body diagram. In this case, the horizontal force applied to block B is missing from the diagram. This force is necessary to overcome the frictional forces and maintain a constant speed. Additionally, the weight of block A should also be included in the forces acting on block B. Therefore, the correct equation for Fnet should be Fnet = F - FfG - FfA - W, where F is the horizontal force applied to block B and W is the weight of block A. Solving for F, we get F = FfG + FfA + W. Plugging in the given values, the magnitude of the horizontal force required to drag block B at constant speed is 5.04 N. It is also important to note that the force applied to block B must be greater than the force of friction in order to overcome it and maintain a constant speed.