Question #185542

The coefficient of kinetic friction between the

tow truck’s wheel and the incline (300) is 0.40.

a. What must be the minimum mass of

the tow truck to start lifting

(accelerating at the rate of 2m/s2) the

10,000N car? Consider the rope has

negligible mass.

b. What values of the tow truck’s mass

will keep the system moving at a

constant speed?


Expert's answer

1)


W+(m+M)a=Mg(sin⁡30−μcos⁡30)10000+100009.82+2M=(9.8)(sin⁡30−0.4cos⁡30)MM=24000 kgW+(m+M)a=Mg(\sin{30}-\mu \cos{30})\\ 10000+\frac{10000}{9.8}2+2M\\=(9.8)(\sin{30}-0.4 \cos{30})M\\M=24000\ kg

2)


W=Mg(sin⁡30−μcos⁡30)10000=(9.8)(sin⁡30−0.4cos⁡30)MM=6600 kgW=Mg(\sin{30}-\mu \cos{30})\\ 10000=(9.8)(\sin{30}-0.4 \cos{30})M\\M=6600\ kg


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