Question

A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to slide it along the floor by pushing horizontally on the side of the refrigerator. The coefficient of static friction between the floor and the refrigerator is 0.30. Depending on where you push, the refrigerator may start to tip over before it starts to slide along the floor. What is the highest distance above the floor that you can push the refrigerator so that it will not tip before it begins to slide?
A.1.2 m

B.1.0 m

C.1.6 m

D. 1.4 m

E. 0.71 m

Answer 1

To solve this problem it is necessary to apply the concepts related to Force of Friction and Torque given by the kinematic equations of motion.

The frictional force by definition is given by

$F= \mu mg$

Our values are here,

$\mu=0.3$

$m=120kg$

$g=9.8m/s^2$

Replacing,

$F=0.30*120*9.8 = 352.8N$

Consider the center of mass of the body half its distance from the floor, that is d = 0.85 / 2 = 0.425m. The torque about the lower farther corner of the refrigerator should be zero to get the maximum distance, then

$F*x = mg*d$

Re-arrange for x,

$x= \frac{mg*d}{F}$

$x= \frac{mg*d}{\mu mg}$

$x= \frac{d}{\mu}$

$x= \frac{0.425}{0.3}$

$x = 1.42m$

Then we can conclude that 1.42m is the distance traveled before turning.