Fridge on Truck Homework: Find Normal Force, Forces Exerted & Rope Tension

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SUMMARY

The discussion focuses on calculating forces acting on a refrigerator during various scenarios on a truck. The normal force exerted on the refrigerator by the truck bed is determined to be 1274 N when the truck is not accelerating. When the truck accelerates at 3 m/s², the horizontal force on the refrigerator is calculated as 390 N. The maximum acceleration before tipping occurs is derived using torque equations, and the tension in the rope when the truck accelerates at twice this maximum is also analyzed.

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Homework Statement



A refrigerator is approximately a uniform parallelepiped h = 8 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 130 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 *
1274 OK
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 *
390 OK
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.

Homework Equations





The Attempt at a Solution



for c) Στ=0 for the fridge to not fall
Στ=m*a*R*h/w-m*g*R*w/h
a=g
a=9.81 m/s^2

this is wrong

for d) i believe i have to set my answer up like this
Στ=0 for the fridge to not fall
Στ=m*a*R*h/w-m*g*R*w/h-T*2*R*h/w
 
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c)

g/h=a/w

If you draw a sketch of the body it's evident. The total acceleration (g and a) vector points to an edge.
 

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