(A) if half of the weight of a small 1.00X10^3 kg utility truck is supported by

Question

(A) if half of the weight of a small 1.00X10^3 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (B) will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (C) solve both problems assuming the truck has four-wheeled drive (A) if half of the weight of a small 1.00X10^3 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (B) will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (C) solve both problems assuming the truck has four-wheeled drive

Explanation / Answer

(A) N = W/2 = (1 x 10^3 x 9.8 / 2) = 4900 N

coefficient of friction on dry concrete =0.7

then f = us N = 0.7 x 4900

a = f/ m = 0.7 x 4900 / 1000 = 3.43 m/s^2

(B) for metal surface , u = 0.6

then a_max = 0.60 x 9.8 = 5.9 m/s^2

so cabinet will not slip.

(C) now N = (1 x 10^3 x 9.8) = 9800 N

a =f/m = 0.7 x 9800 / 1000 = 6.86 m/s^2 ...Ans

now that is more than a_max.

hence cabinet will slide.

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