Engineering Mechanics: Dynamics
8th Edition
ISBN: 9781118885840
Author: James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher: WILEY
Question
Chapter 3.15, Problem 301RP
To determine

## The coefficient of kinetic friction between the truck bed and the crate.

The coefficient of kinetic friction between the truck bed and the crate is 0.382.

### Explanation of Solution

Draw the free body diagram of the crate as shown below.

Write the equilibrium condition along the y-direction.

Fy=0NWcosθ=0N=Wcosθ                                                                                            (I)

Here, the normal force exerted by the truck bed is N, the weight of the crate is W and the angle of inclination of the truck bed is θ.

Write the energy balance equation.

Urel=ΔTrel(WsinθNμk)x=12m(v22v12)(mgsinθNμk)x=12m(v22v12)                                                           (II)

Here, the mass of the crate is m, the acceleration due to gravity is g, the coefficient of kinetic friction is μk, the distance slid by the crate is x, the initial velocity of the crate is v1 and the final velocity of the crate is v2.

Conclusion:

Substitute, Equation (I) in Equation (II).

(mgsinθWcosθμk)x=12m(v22v12)(mgsinθmgcosθμk)x=12m(v22v12)g(sinθcosθμk)x=12(v22v12)                                               (III)

Calculate θ as follows:

tanθ=15100θ=tan1(0.15)θ=8.53°

Substitute, 9.81m/s2 for g, 8.53° for θ, 2 m for x, 0 for v2 and 3m/s for v1 in Equation (III).

(9.81m/s2)(sin8.53°μkcos8.53°)(2 m)=12[02(3m/s)2](sin8.53°μkcos8.53°)=12[(3m/s)2(2 m)(9.81m/s2)]0.14830.9889μk=0.22940.9889μk=0.1483+0.2294μk=0.37770.9889μk=0.382

Thus, the coefficient of kinetic friction between the truck bed and the crate is 0.382.

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