Determine the horizontal force P required to cause slippage

Determine the horizontal force P required to cause slippage to occur. The friction coefficients for the three pairs of mating surface are indicated. The top block is free to move vertically. Answer: P = n the tolerance is +/-2%

Solution

<----------- F1

-------------------> P

<--------------- F2

<---------------- F3

Let the Frictional force from top to bottom be as F1 , F2 , F3.

the frictional force F = frictional coefficient * normal force.

where normal force = weight of the body above frictional surface.

therefore F1 = 0.49 * 105 * 9.81 = 504.7245 N

F2 = 0.38 * (105+50) * 9.81 = 577.809 N

F3 = 0.25 * (105+50+31) * 9.81 = 456.165 N.

F2 > F1 > F3.

When P is equal to F3 (= 456.165 N ) the system will never move because the frictional force at the interface of 105 kg and 50 kg is higher than F3 and the mass 105 kg is constrained in the horizontal direction.

when the P = F1 + F3 ( 504.7245 + 456.165 =960.8895 N) , the system of 31kg and 50 kg starts sliding underneath the weight of 105 kg as a single system since (F2+F1) > (F1+F3).

Therefore the horizontal force P required for the system under 105 kg to slip over the base surface is 960.8895 N.