A mass of 7 kg is placed on a horizontal table with friction. A rope is tied to

Question

A mass of 7 kg is placed on a horizontal table with friction. A rope is tied to it and it is slung over a pulley and the other end of the rope is attached to a hanging mass of 19 kg. The hanging mass is great enough suchthat it overcomes the static friction of the mass on the table and the hanging mass accelerates downward at 2.9 m/s^ . If the coefficient of static friction is 3 times larger than the coefficient of kinetic friction, and if the hanging mass is changed, what is the minimum the hanging mass needs to be to begin to move the mass on the table, in kg?

Explanation / Answer

m2g - T = m2a

T- um1g = m1a ( in case of motion kinetic friction will act)

adding both

(m2g-um1g)/(m1+m2) = a

putting values to calculate u ( kinetic friction coefficient)

19*9.8 - u7*9.8 = 26*2.9

=> u kinetic = 1.61

u static = 3*1.61 = 4.83

in case of no motion staic friction will act and a=0

=> m2g= um1g

=> m2 = 4.83*7

=> m2 = 33.81 kg

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