"2 Block Table Friction Problem

[KSimmonsUGA]

Homework Statement

A 3.5 kg block (block A) sits on a 6.2 kg block (block B). The larger block is sitting on a horizontal frictionless table. The coefficients of friction between the blocks are static = 0.35 and kinetic = 0.22. a) Calculate the maximum horizontal force that can be applied to teh larger blcok if the smaller block is not to slide on the larger block.
b) For the force found in part a, calculate the acceleration of each block.
c) A 35 N force is exerted on the top block at an angle of 20 degrees below the horizontal. The force from part A has been removed. Calculate the acceleration of each block.

Homework Equations

Sum of Forces = m*a

The Attempt at a Solution

Okay, I know I need to start with a free body diagram, but I am unsure that even that is correct. So far, I have:

Table: weight (a+b) acts down on the table, while table pushes up with normal force (Fnormal) equal to the weight of (a+b).

Block B: Weight (a+b) pushing down while the Fnormal (a+b) pushes up. Force(kinetic) pushes to the right, while Force(static) pushes to the left.

Block A: Weight (a only) pushing down while Fnormal (a+b) pushes up. Force(kinetic) pushes to right, while Force(static) pushes to the left.

 

[jbsteele]

a) I think this one is simple enough. The maximum horizontal force that can be applied to the larger block is equal to the static coefficient of friction times the weight of the smaller block, or 0.35*3.5 = 1.225 N.b) a) Again, I think this is fairly simple. Fnet = m*a. For the larger block, Fnet = (6.2)*a and for the smaller block Fnet=(3.5)*a. So, a = Fnet/m. For the larger block, a = 1.225/6.2 = 0.198 m/s^2 and for the smaller block, a=1.225/3.5 = 0.347 m/s^2. Is this correct?c) For the larger block, Fnet = 35cos(20) - 0.22*3.5 = 28.48N. So, a = 28.48/6.2 = 4.59 m/s^2. For the smaller block, Fnet = 35cos(20) + 28.48 - 0.35*6.2 = 15.39N. So, a = 15.39/3.5 = 4.41 m/s^2. Is this correct?

 

[baba89]

As a scientist, the first step in solving this problem would be to identify the key components and variables involved. In this case, we have two blocks (A and B), a table, and various forces acting on the blocks.

Next, we can use the given information to create a free body diagram for each block. This will help us visualize the forces acting on each block and their directions.

For block A, we have the weight (mg) acting downwards, the normal force (Fnormal) acting upwards from the table, and the frictional forces (Fstatic and Fkinetic) acting in opposite directions horizontally.

For block B, we have the weight of both blocks (mg) acting downwards, the normal force (Fnormal) acting upwards from the table, and the frictional forces (Fstatic and Fkinetic) acting in opposite directions horizontally.

Now, using the equations of motion (sum of forces = mass x acceleration), we can solve for the maximum horizontal force that can be applied to block B without causing block A to slide. This force would be equal to the static frictional force (Fstatic) acting on block A.

For part b, we can use the same equations to calculate the acceleration of each block with the maximum applied force. This would give us the acceleration of block B (since it is the one being pushed) and the acceleration of block A (which would be moving with block B).

Finally, for part c, we can use the given information to calculate the net force acting on block B (from the 35 N force at an angle of 20 degrees) and then use the equations of motion to calculate the acceleration of each block in this scenario.

Overall, the key to solving this problem is to accurately identify the forces acting on each block and use the appropriate equations to calculate the desired quantities.