If the weight on the driving wheels of a tractor is 93,800 lb and it is moving on firm earth with a coefficient of traction of 0.55, what is the usable power if the tractor is operating in Abha (elevation of approx. 7,000 ft above sea level)?

A. 51,590 lb
B. 55,440 lb
C. 40,760 lb
D. 45,400 lb

Explanation: In order to find out the usable power, we first calculate the traction force. The traction force is the maximum force that the tractor can exert. It is equal to the product of the normal force and the coefficient of traction. The normal force is equal to the weight on the driving wheels, which is given to be 93,800 lb.The coefficient of traction is 0.55. So, the traction force is: 93,800 × 0.55 = 51,590 lbNow, we use the formula for usable power:Usable power = Traction force × Speed ÷ 375The speed is not given, so we cannot calculate the usable power. The usable power depends on both the traction force and the speed. A higher speed will require a higher usable power. A lower speed will require a lower usable power.

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If the weight on the driving wheels of a tractor is 93,800 lb and it is moving on firm earth with a coefficient of traction of 0.55, what is the usable power if the tractor is operating in Abha (elevation of approx. 7,000 ft above sea level)?A. 51,590 lb B. 55,440 lb C. 40,760 lb D. 45,400 lb

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If the weight on the driving wheels of a tractor is 93,800 lb and it is moving on firm earth with a coefficient of traction of 0.55, what is the usable power if the tractor is operating in Abha (elevation of approx. 7,000 ft above sea level)?

A. 51,590 lb
B. 55,440 lb
C. 40,760 lb
D. 45,400 lb