Determine the force:

Determine the force P needed to start block B illustrated in Figure moving to the right if the coefficient of friction is 0.3 for all surfaces of contact. Block A weighs 80 N and block B weighs 160 N.

Solution

Figure illustrates the free body diagrams for block A and block B. As the possible motion of the block B is towards right, the direction of frictional forces F_A & F_B are illustrated acting towards left, for block B. While block B moves towards right the relative motion of block A shall be towards left and therefore the frictional force FA for block A is illustrated acting towards right.

Now the conditions of equilibrium may be applied to determine the force P.

For Block A

                                        ∑ F_x  = 0

                             ∴ F_A  = T cos 30^o  = 0.866 T                                       ------------ (1)

 (where, T is the tension in the given cord)

                                            ∑ F_y  = 0

                                  ∴ N _A = W_A  - T sin 30^o

                                              = 80 - 0.5 T                        -------------- (2)

Also                           F_A  = μ N _A

                     ∴ 0.866 T = 0.3 (80 - 0.5 T)

∴ T = 24/1.016            = 23.62 N                             -------- (3)

 For Block B

∴ F_A  = 0.866 × 23.62 = 20.45 N

∴ N _A  = 80 - 0.5 × 23.62 N

= 68.19 N

∑ F_y  = 0

∴ N _B  = N _A  + W_B

= 68.19 + 160

= 228.19 N                 ---------- (4)

F_B  = μ N B

= 0.3 × 228.19

= 68.46 N

∑ F_x  = 0                   ----------------( 5)

∴ P = F_A  + F_B

= 20.45 + 68.46

= 88.91 N                    ----------- (6)

Thus, a force 88.91 N is needed to move the block B to the right.