In Fig. 3-28, how large a force F is needed to give the blocks an acceleration of 3.0
m/s2 if the coefficient of kinetic friction between blocks and table is 0.20? How large
a force does the 1.50-kg block then exert on the 2.0-kg block?
The maximum force
Answer to Problem 79SP
Solution:
Explanation of Solution
Given data:
Arrangement of the blocks of mass
Coefficient of friction between blocks and table is
The needed acceleration of the blocksis
Formula used:
Write the expression for the first condition of the force’s equilibrium:
Here,
From the Newton’ s second law of the motion, the expression of the force is
Here,
Write the expression for the friction force:
Here,
Sign convention: Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Also, consider the leftward force as negative and rightward forces are positive.
Explanation:
Draw the free body diagram of a
In
Recall the expression for the first condition of the force’s equilibrium:
From the free body diagram:
Substitute
Recall the expression for the friction force:
Substitute
Recall the expression for Newton’s second law of the motion:
From free body diagram, the force equation is
Substitute
Draw the free body diagram of the all the blocks:
In
Now, consider the
From free body diagram:
Substitute
Recall the expression of the friction force:
Substitute
Recall the expression for Newton’s second law of the motion:
From free body diagram:
Substitute
Draw the free body diagram of the all the blocks:
In the above diagram,
Now, consider the
From free body diagram:
Substitute
Recall the expression for the friction force:
Substitute
Recall the expression for Newton’s second law of the motion:
From free body diagram:
Substitute
Conclusion:
Therefore, the required force
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Chapter 3 Solutions
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
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