Question

Block B in the figure below weighs 704 N. The coefficient of static friction between block and table it 0.24; angle theta is 28 degree; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. N

Answer 1

∑F = 0 = T₁ - µn
T₁ = µn------>equation (1)

The vertical forces on w₂ :

∑F = 0 = T₂ - w₂
T₂ = w₂------>equation (2)

Around the "knot" (horizontally and vertically) :

∑F = 0 = TcosѲ - T₁
TcosѲ = T₁----->equation (3)

∑F = 0 = TsinѲ - T₂
TsinѲ = T₂------->equation (4)

Recall the trig identity that tanѲ = sinѲ / cosѲ, and divide (4) by (3). You get :

tanѲ = T₂ / T₁
tanѲ = w₂ / µn

Since the block on the table is perfectly horizontal, the normal force (n) is equal to its weight, or 704N. Therefore :

tan 28° = w₂ / (0.24 x 704N)
w₂ = tan 28°(0.24 x 704N)
= 89.8376N