Question

The horizontal position of the 500-kg rectangular block of concrete is adjusted by the 5o
wedge under the action of the force P. If the coefficient of static friction for both wedge
surfaces is 0.30 and if the coefficient of static friction between the block and the horizontal
surface is 0.60, determine the least force P required to move the block. Neglect the weight of
the wedge

Answer 1

Given: Weight of concrete block is 500 kg, inclination is 5°, coefficient of static friction of wedge surfaces is 0.30 and of static friction between the block and the horizontal surface is 0.60.

To find: Force P required to move the block?

Solution:

• Now, lets define and consider omega as the coefficient of friction between the block and the wedge and ∅ as the inclination of wedge.

• Now, we know the formula of omega as ratio of coefficient of friction between (the block and the ground) to the  coefficient of friction (between the wedge surface)

• So putting the values in the formula we get:

           omega = 0.3/0.6

           omega = 0.5

• Now, lets find the forces which are acting on the block:

           weight = mg

           mg = 500 x 9.82

                 = 4910 N

           frictional force F = (omega) x mg

           F = 0.5 x 4910

              = 2455 N

• Now, resolving the forces along the axes, we get:

           mg sin∅ + P = F

            P = F - mg sin∅

            P = 2455 - 4910 sin(5°)

               = 2455 - 429.1

               = 2025.9 N

Answer:

       So, the least force P required to move the block is 2025.9 N.