A slab of weight 40 N rests on a horizontal floor. A block of weight WB = 80 N rests on top of the slab. The coefficient of friction between the block and the slab is μs = 1/3. The coefficient of friction between the slab and the table is μk = 1/4. A horizontal external force F is applied on the slab. What is the maximum value of F so that there is still no relative motion between the block and slab?

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter6: Energy Of A System
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A slab of weight 40 N rests on a horizontal floor. A block of weight WB = 80 N rests on top of the slab. The coefficient of friction between the block and the slab is μs = 1/3. The coefficient of friction between the slab and the table is μk = 1/4. A horizontal external force F is applied on the slab. What is the maximum value of F so that there is still no relative motion between the block and slab? (Show work)

 

Expert Solution
Step 1

Given:

Weight of slab (Ws) = 40 N.

Weight of block (Wb) = 80 N.

Coefficient of friction between block and slab (µs) = 0.33

Coefficient of friction between slab and table (µk) = 0.25

Physics homework question answer, step 1, image 1

Step 2

The free body diagram of block and slab is given as,

Physics homework question answer, step 2, image 1

Physics homework question answer, step 2, image 2

Step 3

The equilibrium equation of box,

Physics homework question answer, step 3, image 1

Acceleration of block due to applied force,

Physics homework question answer, step 3, image 2

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