The cabinet in the figure below weighs 400 N and has width w = 58.0 cm and heigh

Question

The cabinet in the figure below weighs 400 N and has width w = 58.0 cm and height l = 170 cm. A force F is applied horizontally at the upper edge. (a)_What is the minimum force required to start to tip the cabinet? (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. Your response differs from the correct answer by more than 100%. direction 75.25 above the horizontal (applied at the upper left corner)

Explanation / Answer

F*costheta*L + F*sinhtheta*L = Fg*w/2

F*( costheta + sintheta )*1.7 = Fg/2 = 400*0.58/2 = 68.235 N

F = 68.235/(costheta+sintheta)

for F to be minimum , (costheta+sintheta) is maximum .the value of (costheta+sintheta) is maximum for theta = 45

F = 48.25 N

dirction = 45 degrees

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