Question

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

Answer 1

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.  

Answer 2

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!