How Do Tractive and Braking Forces Affect Vehicle Motion on Different Gradients?

[paul7776]

tractive motion, with gradients, Hard questions!

4.1 An engine mass of 95 tonnes is hauling a train of mass 750 tonnes along a section of level track. the tractive resistance to motion of the engine is 120 N/tonne and the rolling resistance for the train is 90 N/tonne. initially the engine is traveling at 27 km/hr and then it accelerates over a distance of 1500m.

A) Calculate the power required to achieve this increase in speed?
B)The train then meets a downgrade of 1V in 110H and is brought to a stop in a distance of 900m. calculate the braking force required to stop the train?

4.7 A large truck mass of 35 tonnes is traveling down a gradient of 1V in 60H at a speed of 90 km/hr. The rolling resitance to motion is 115 N tonne.

A) If when the brakes are applied the truck is brought to a stop in 400m what is the applied braking force?
B)How much power is required to provide this braking force?
C)If this same power were available on a uphill gradient of 1 in 60 what constant speed would the truck be capable of reaching?

If u cannot convert km/hr in2 metres per second (ms-1) its (km/hr X 1000) /3600

 

[tiny-tim]

Welcome to PF!

Hi paul7776! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[nlightner]

A) To calculate the power required to achieve the increase in speed, we can use the formula P = Fv, where P is power in watts, F is force in newtons, and v is velocity in meters per second. First, we need to convert the initial velocity of 27 km/hr to meters per second by multiplying it by 1000 and dividing by 3600. This gives us a velocity of 7.5 m/s. Then, we can calculate the total tractive resistance to motion by multiplying the engine mass (95 tonnes) by the tractive resistance (120 N/tonne), giving us a force of 11,400 N. The rolling resistance for the train is 90 N/tonne, so we multiply the train mass (750 tonnes) by this resistance, giving us a force of 67,500 N. The total resistance to motion is then 79,900 N. So, the power required to achieve the increase in speed is P = (79,900 N)(7.5 m/s) = 599,250 watts.

B) To calculate the braking force required to stop the train, we can use the formula F = ma, where F is force in newtons, m is mass in kilograms, and a is acceleration in meters per second squared. First, we need to convert the distance of 900m to meters by simply leaving it as 900m. Then, we can calculate the acceleration by dividing the final velocity (0 m/s) by the time it takes to stop (unknown). So, our formula becomes F = (750,000 kg)(0 m/s)/t. We can cancel out the mass and velocity, leaving us with F = 0/t. Since the train is brought to a stop, the time it takes is 0 seconds. This means that the braking force required is 0 newtons.

A) To calculate the applied braking force, we can use the same formula as in part B, F = ma. However, we now have a distance and a final velocity, so we can solve for the acceleration and then use that to find the force. First, we need to convert the speed of 90 km/hr to meters per second by multiplying it by 1000 and dividing by 3600. This gives us a velocity of 25 m/s. Then, we can calculate the acceleration by using the formula a = (vf - vi)/t,