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A 140 N block rests on a table. The suspended mass has a weight of 77 N.

AWhat is the magnitude of the minimum force of static friction required to hold both blocks at rest?

Answer in units of N.

BWhat minimum coefficient of static friction is required to ensure that both blocks remain at rest?

Verified Answer :

ekal

I'm assuming that this is what your diagram looks like because these types of problems always look like this (lol).

Comment if you need clarification!

View image ekal
asifejaz20

Answer:

a) Force required to keep both the masses in rest will be - 77 N  

b) The coefficient of friction will be 0.55

Explanation:

a) As the weight of 77 N is suspended vertically so, there must act 77 N force on the 140 N weight to keep both masses in rest. Hence, the answer is -77 N.

b) We know that:

                            140µ = 77

                                  µ = 77/140

                                  µ = 0.55

Hence, the minimum co-efficient of friction required to keep both bodies in rest will be 0.55.  

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