Question

A 3.06kg wood block is pressed against a vertical wood wall by a force applied at angle ϕ=32.6° to the horizontal. If the block is initially at rest, and μs=0.313, what force would you need to apply to begin moving the box up the wall?

Answer 1

Let applied force against the wall is F

now the Normal force on the block is given by

$F_n = F cos32.6$

now friction force on the block will be given as

$F_f = \mu * F_n$

$F_f = 0.313* (F cos32.6)$

now net downwards force on the block will be

$F_d= F_f + mg$

$F_d = 0.313*(Fcos32.6) + 3.06*9.8$

$F_d = 0.264F + 29.99$

now this net downward force must be counterbalanced by upwards applied force

$F_u = F_d$

$Fsin32.6 = 0.264F + 29.99$

$0.275*F = 29.99$

$F = 109.1 N$

so it required 109.1 N force to move it upwards