Rimpull/Rolling resistance

[talaroue]

Homework Statement

1.A Ferrari has a V-12 engine rated at 540 hp, it has a maximum speed of 196 mph in sixth gear if the engine efficiency is 90%. Determine the maximum rimpull. The Ferrari weighs 4,055 lb and is operated over a concrete road with rolling resistance of 40 lb/ton and a slop of 3%. Determin the maximum enxternal pull the Ferrari has when operated in sixth gear at maximum speed.

Homework Equations

Rimpull=375*hp*effiency/mph

The Attempt at a Solution

I simply plugged and chugged and got 928.8 lb
=375*540*.9/196

this is wrong i believe because I think I have to incorporate the slope in it but I am not sure how. It is an extra credit problem so its not in my book.

 

[anathema]

Hello,

I would approach this problem by first breaking it down into smaller steps and using the relevant equations.

Step 1: Determine the force required to overcome rolling resistance
The equation for rolling resistance is Frr = Crr * W, where Frr is the force required to overcome rolling resistance, Crr is the coefficient of rolling resistance, and W is the weight of the Ferrari. We are given that Crr = 40 lb/ton and W = 4,055 lb. Converting 40 lb/ton to lb/lb, we get Crr = 0.02 lb/lb. Plugging in these values, we get Frr = 0.02 * 4,055 = 81.1 lb.

Step 2: Determine the force required to overcome the slope
The equation for the force required to overcome a slope is Fsl = W * sin(theta), where Fsl is the force required to overcome the slope, W is the weight of the Ferrari, and theta is the slope angle. We are given that the slope is 3%, which can also be written as 0.03. Plugging in these values, we get Fsl = 4,055 * sin(0.03) = 6.1 lb.

Step 3: Determine the force required to maintain maximum speed
The equation for the force required to maintain maximum speed is Fmax = (W * V^2) / (2 * g * eta), where Fmax is the force required to maintain maximum speed, W is the weight of the Ferrari, V is the maximum speed, g is the acceleration due to gravity (9.8 m/s^2), and eta is the efficiency of the engine. We are given that V = 196 mph and eta = 0.9. Converting mph to m/s, we get V = 87.44 m/s. Plugging in these values, we get Fmax = (4,055 * (87.44)^2) / (2 * 9.8 * 0.9) = 1,727,671.6 lb.

Step 4: Determine the maximum external pull
The maximum external pull is the sum of the forces required to overcome rolling resistance, slope, and maintain maximum speed. Therefore, the maximum external pull is Fmax + Frr + Fsl = 1,727,671.6 +

 

[tomcue]

I would first clarify the definitions of rimpull and rolling resistance. Rimpull is the force that a vehicle's engine can produce at its wheels, while rolling resistance is the force that opposes the motion of a vehicle when it is in motion on a surface. It is important to note that these forces are not constant and can vary depending on various factors such as speed, weight, and surface conditions.

To calculate the maximum rimpull, we can use the formula Rimpull=375*hp*efficiency/mph. Plugging in the given values, we get Rimpull=375*540*0.9/196= 928.8 lb. This means that the maximum force that the Ferrari's engine can produce at its wheels is 928.8 lb.

To calculate the maximum external pull, we need to consider the rolling resistance and the slope of the road. The maximum external pull will be the sum of the rimpull and the rolling resistance, taking into account the slope of the road. In this case, the rolling resistance is given as 40 lb/ton and the slope is 3%. To convert the weight of the Ferrari from pounds to tons, we divide by 2000, giving us 2.0275 tons. Thus, the rolling resistance can be calculated as 40*2.0275= 81.1 lb.

To incorporate the slope, we can use the formula F=ma, where F is the force, m is the mass, and a is the acceleration. In this case, we are looking for the maximum force (F) and we know the mass (m) and the acceleration (a) due to the slope. The acceleration can be calculated as a=gsinθ, where g is the gravitational acceleration (9.8 m/s^2) and θ is the slope in radians. Converting 3% to radians, we get θ=0.03*2π= 0.1885 radians. Thus, the acceleration can be calculated as a= 9.8*sin(0.1885)= 1.8 m/s^2.

Plugging in the values in the formula F=ma, we get F= 4.055*1.8= 7.3 lb. Therefore, the maximum external pull can be calculated as 928.8+81.1+7.3= 1017