The cabinet in the figure below weighs 400 N and has width w = 61.0 cm and heigh
ID: 1461823 • Letter: T
Question
The cabinet in the figure below weighs 400 N and has width w = 61.0 cm and height = 110 cm. A force F with arrow 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? (c) Find the magnitude and direction of the minimum force required to tip the cabinet if the point of application can be chosen anywhere on the cabinet. 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*0.61/(2*1.1)
= 110.9 N
b) before sliding, Fnet = 0
F - mue_s*N = 0
mue_s*N = F
mue_s = F/N
= 110.9/400
= 0.277
c) angle made by diagonal with horizontal, theta = tan^-1(l/w)
= tan^-1(110/61)
= 61 degrees
when we apply force at upper left corner at 29 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*0.61/(2*sqrt(1.1^2 + 0.61^2))
= 97 N at 29 degrees above horizontal
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