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 8,000 kg load sits on the flat bed of a 15,000 kg truck moving at 10.0 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. ____ m

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

_ mass of load

_ speed of truck

_ mass of truck

_ coefficient of friction

Answer 1

Speed of truck , Initial Velocity = 10m/s
Let minimum stopping distance = d
Final Velocity = 0
Therefore,
V^2 = u^2 - 2*a*s
2*a*d = 10^2 --------1

Total Friction Force = u*N = 0.475 * m*g
We know, F = m*a

Friction Force = M*a
0.475 * m*g = m * a

a = 4.655 m/s^2

Substituing Value in 1
2 * 4.655* d = 100

d = 100/(2*4.655)

d = 10.74 m

Minimum Stopping Distance d = 10.74 m

Both the masses are not necessary for calculation.