The cabinet in the figure below weighs 400 N and has width w = 51.0 cm and heigh
ID: 1461757 • Letter: T
Question
The cabinet in the figure below weighs 400 N and has width w = 51.0 cm and height ? = 120 cm. A force is applied horizontally at the upper edge.
(a) What is the minimum force required to start to tip the cabinet? N
(b) What is the minimum coefficient of static friction required for the cabinet not to slide with the application of a force of this magnitude?
magnitude
N
direction
° above the horizontal (applied at the upper left corner)
magnitude
N
direction
° above the horizontal (applied at the upper left corner)
Explanation / Answer
a) Apply, before tipping Net torque about right edge = 0
400*(w/2) - F*l = 0
F*l = 400*W/2
F = 400*w/(2*l)
= 400*0.51/(2*1.2)
= 85 N
b) before sliding, Fnet = 0
F - mue_s*N = 0
mue_s*N = F
mue_s = F/N
= 85/400
= 0.2125
c) angle made by diagonal with horizontal, theta = tan^-1(l/w)
= tan^-1(120/51)
= 67 degrees
when we apply force at upper left corner at 23 degrees with horizontal the cabinte can be tipped easily.
Apply, Net torque about bottom right corner = 0
400*w/2 - F*sqrt(l^2 + w^2) = 0
F = 400*w/(2*sqrt(l^2+w^2)
= 400*0.51/(2*sqrt(1.2^2 + 0.51^2))
= 78 N at 23 degrees above horizontal
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