While hauling a log in the back of a flatbed truck, you are pulled over by the s

Question

While hauling a log in the back of a flatbed truck, you are pulled over by the state police. Although the log can't roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. You claim that the truck couldn't possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now your day in court is approaching.

1)The log has a mass of m = 844 kg; the truck has a mass of M = 9750 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 26.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction ?s needed to keep the log in the back of the truck.

Explanation / Answer

Change in speed = 55mph = 24.58 m/sec

Acceleration = 24.58/26 = 0.945 m/sec^2

Force that engine is capable of applying(calculated for empty truck) = ma = 9750*0.945 = 9218.23N

This is the maximum capability of engine which will remain constant for any condition.

When log of wood is loaded

Combined mass = 9750+844 = 10594 Kg

Maximum possible acceleration = F/m = 9218.23/10594 = 0.87 m/sec^2

For this magnitude of acceleration for the block, frictional force on the log of wood should be = mass of wood * acc

= 844*0.87 = 734.39N

This force = static friction coefficient * N = mu * m * g

So, mu = F/(mg) = 734.39 / (844*9.81) = 0.088

This is the minimum required static coefficient of friction.

I hope you got this.

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