The acceleration is 3.08 m/s².
The Center of gravity of the refrigerator is 78.0 cm = 78*10^-2 m above the midpoint of the base.
The center of gravity distance from either edge of the refrigerator is =56 /2 =28 cm =28*10^-2 m.
Let acceleration of the truck is = a
And the mass of the truck is = m
Weight of the refrigerator =m*g
The horizontal force on the refrigerator is F = ma
Torque on center of gravity due to F is T1 = F × 78 × 10^-2 Nm =m×a×81×10^-2
The torque acting on the center of gravity due to the weight of the refrigerator is T2 = m*g*28*10^-2
for the refrigerator does over T1=T2
m*a*78*10^-2 = m*g*28*10^-2
The maximum acceleration the truck can have so that the refrigerator does over
a = g*28/78
= 3.08 m/s^2
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