
Concept explainers
(a)
To Calculate: The maximum and minimum values of applied force for which the block does not slip.
(a)

Answer to Problem 63P
Explanation of Solution
Given data:
Mass of the block,
Mass of the wedge,
Coefficient of static friction,
Angle of wedge,
Formula Used:
Newton’s second law of motion:
Where, m is the mass and a is the acceleration.
Calculation:
Free-body diagram:
Maximum force,
Minimum force,
Where,
For the block,
Substitute (3) in (1) to find the minimum acceleration:
Substitute the values and solve:
For maximum force, reverse the direction of f
Substitute the values and solve:
Conclusion:
The maximum and minimum values of applied force for which the block does not slip are 84 N and 1.6 N respectively.
(b)
To Calculate: The maximum and minimum values of applied force for which the block does not slip.
(b)

Answer to Problem 63P
Explanation of Solution
Given data:
Mass of the block,
Mass of the wedge,
Coefficient of static friction,
Angle of wedge,
Formula Used:
From previous part:
Calculation:
Substitute the values and solve for minimum force:
For maximum force, reverse the direction of f
Substitute the values and solve:
Conclusion:
The maximum and minimum values of applied force for which the block does not slip are 37.5 N and 5.8 N respectively.
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