   Chapter 4, Problem 43P

Chapter
Section
Textbook Problem

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, a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?

(a)

To determine
The minimum stopping distance.

Explanation

Given Info: Mass of the load is 10000 kg. Mass of the truck is 20000 kg. Velocity of the truck is 20 m/s. The co-efficient of static friction ( μs ) is 0.500.

The free body diagram is given below.

From the diagram,

N=mlg (I)

Formula for the force of static friction is,

Fs=mla (II)

• Fs is the force of static friction.
• N is the normal force.
• ml is the load mass.

Negative sign because the truck is slowing down which causes the acceleration to be negative.

The force of static friction is,

Fs=μsN (III)

From Equations (I), (II) and (III),

mla=μsmlg

The acceleration is,

a=μsg

From Newton’s equations of motion, the stopping distance is,

s=vf2vi22a

• vf is the final velocity

(b)

To determine
The minimum stopping distance.

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