A refrigerator is approximately a uniform parallelepiped h = 7 ft tall, w = 3 ft

Question

A refrigerator is approximately a uniform parallelepiped h = 7 ft tall, w = 3 ft wide, and d = 2 ft deep. It sits upright on a truck with its 3 ft dimension in the direction of travel. Assume that the refrigerator cannot slide on the truck and that its mass is 140 kg. For the first three parts of this problem, the rope shown in the picture is not there.

a) When the truck is not accelerating, what is the normal force exerted on the refrigerator by the truck bed?
N = N

b) If the truck now accelerates at 3 m/s2, what is the horizontal force exerted on the refrigerator by the truck bed?
f = N

c) What is the maximum acceleration the truck can have such that the refrigerator does not tip over?
amax = m/s2

d) Suppose now that a rope connects the top of the refrigerator with the cab of the truck, which now accelerates at twice the acceleration calculated in (c). The refrigerator lifts off slightly at the front but is held in place by the horizontal rope. Find the tension in the rope.
T = N

Explanation / Answer

a) N=140*g 1372 N b) F=m*a F=140*3 420 N c) sum torque about the back edge of the fridge St=0 for the fridge to not fall St=m*a*R*h/w-m*g*R*w/h Where R is the distance from the back edge to center of mass of the fridge. a=g a=9.81 m/s^2 d) again sum torque about the back edge of the fridge St=0 for the fridge to not fall St=m*a*R*h/w-m*g*R*w/h-T*2*R*h/w Where R is the distance from the back edge to center of mass of the fridge. a=2*g Solve for T T=m*g*(2-w^2/h^2)/2 plug in the numbers T=140*9.81*(2-3^2/7^2)/2 T=1247.27N

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