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 11,000 kg load sits on the flat bed of a 25,000 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 0.475 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.
Ff = coefficient of friction x m x g
=>Ff = 0.475 x 11000 x 9.8 = 51205 N
Let the retardation (a) maximum, which can be produced:-
=>By F = ma
=>51205 = (11000+25000) x a
=>a = 1.42 m/s^2
(a) By v^2 = u^2 – 2as
=>0 = (12)^2 – 2 x 1.42 x s
=>s = 50.70m
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