Question

A delivery truck is carrying a 120 kg refrigerator. The refrigerator is 2.20 m tall and 85.0 cm wide. The refrigerator is facing sideways and is prevented from sliding. The center of gravity of the refrigerator is located at its geometrical center. Determine the maximum acceleration that the truck can have before the refrigerator begins to tip over.

Answer 1

Answer:

3.79 m/s²

Explanation:

Given that:

The mass of refrigerator = 120 kg

The height of refrigerator = 2.20 m

The width of refrigerator = 85.0 cm = 0.85 m

Since the center of gravity is located at the geometrical center.

Then;

the center mass of refrigerator = h/2 = 2.20 m / 2

= 1.10 m

From width of refrigerator, the distance of center mass to either edge of refrigerator = 0.85 m / 2

= 0.425 m

So; the torque acting on the center mass due to the force F is:

$\tau_1= F(1.01 \ m)$

the torque acting on the center mass due to the weight of the refrigerator is:

$\tau_2 = w(0.425 \ m)$

Thus, since the refrigerator does not tip over,

$\tau_1 = \tau_2$

F(1.10 m) = w(0.425 m)

ma (1.10 m) = mg (0.425 m)

a (1.10 m) = g(0.425 m)

$a = \dfrac{9.8\times 0.425}{1.10}$

a = 3.79 m/s²