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A 500-N concrete block is to be lifted by the pair of tongs shown.

Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at points F and G.

Answer :

The smallest allowable value of the coefficient of static friction between the block and the tongs at F and G is 0.4.

The maximum force of static friction, Fs, can be calculated using the equation Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force N is equal to the weight of the block, which is given as 500 N.

To determine the smallest allowable value of the coefficient of static friction, we need to find the maximum force of static friction at F and G. Since the tongs are pulling vertically upwards, the normal force at both points F and G will be equal to the weight of the block, which is 500 N.

Substituting these values into the equation Fs ≤ μsN, we get:

Fs ≤ 0.4 × 500

Simplifying the equation, we find:

Fs ≤ 200

Therefore, the maximum force of static friction at F and G is 200 N. This means that the smallest allowable value for the coefficient of static friction is 0.4, in order to prevent the block from slipping when lifted by the tongs.

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