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 11,000-kg load sits on the flatbed of a 20,000-kg truck moving at 11.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.650 with the truck bed.
(a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.

Explanation / Answer

(a) There is no relative motion between the truck and the load if they have the same acceleration. Let their accelerations be a. Then a is in the direction opposite to that of the velocity of the truck (only then can the truck stop). Let the velocity of the truck be towards right and the acceleration be towards left. Now make the free-body diagram for the load : Weight (mg) acts vertically downwards Normal reaction (N) acts vertically upwards So N = mg Force of friction (f) acts towards left (since the block has a tendency to slip towards right) Maximum value of friction , f = uN = umg So f = ma or a = ug = 4 m/s^2 Now consider the truck : Initial velocity, u = 11 m/s Final velocity, v = 0 m/s a = - 4m/s^2 (- sign shows that a is in the direction opposite to that of u) So s = (v^2 - u^2)/2a = 15.125 m (b) Mass of the load and the truck are unnecessary.

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