Consider a large truck carrying a heavy load, such as steel beams. A significant

Question

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 9,000 kg load sits on the flat bed of a 20,000 kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.575 with the flat bed of the truck.

(a) Calculate the minimum stopping distance for the truck for which the load will not slide forward relative to the truck.
??? m

(b) Is any piece of data unnecessary for the solution? (Select all that apply.)

coefficient of friction

speed of truck

mass of truck

mass of load

???

Explanation / Answer

V^2 = U^2 + 2*aS

0 = 12^2 - 2a*S

=> a = 72/S

now

Pseudo Force on the load by this acc = m*72/S

m*72/S = umg ( in case of no slipping)

=> S = 72/ug = 72/(9.8*0.575) = 12.77 m

b) mass of truck is not needed at all

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