Work, power and energy transfer in dynamic engineering systems

[lukeodn]

1. Machanical Prinisiples assignment.

Homework Equations

Hi guys, i am a new member here, and am really looking for some help.

My question is: A wheeled truck has a total mass when loaded of 1500 kg and is to be hauled up an incline of 1 in 1o (sine). g= 9.81 m/s2 and the coefficiant of friction between the wheeles and the incline surface is 0.05, A tractive force is applied from a vehicle through a tie-bar, sufficient to acceleration the truck from rest to a verlocity of 20 m/s in 10s.
Calculate:
a. The tractive force delivered through the tie bar during the acceleration
b. The total work done on the wheeled truck during this period
c. The average power expanded on the truck by the towing vehicle
d. The maximim power developed by the towing vehicle in order to accelerate the truck.


3. After trying these questions i have found my self more confused than what i was originally!

Thankyou for any help, will be much appreciated!

 

[RTW69]

I am unclear about of the slope for this problem so I am assuming the truck is going up a slope of 1 degree. Since the truck goes from 0 to 20 m/s in 10 seconds and the acceleration equals the change in velocity divided by the change in time the acceleration is 2 m/s^2.

The distance the truck travels is .5*a*t^2 or 100 meters.

Now you have to draw a free body diagram of the truck on the slope and sum the forces perpendicular to the slope and parallel to the slope. The sum of the forces perpendicular to the slope are equal to zero and will give you the normal force which is 14,700 Newtons

The normal force is important because the frictional force= Normal force * u_k or 735 Newtons

Now you must sum the forces parallel to the slope. The force needed to accelerate the truck 2 m/s^2 while overcoming the force of gravity acting down the slope plus the frictional rolling force is 3,990 Newtons.

Work is the force times distance. You know both so work is 399,000 N*m

Average power is Force times change in distance divided by change in time. You know all three so average power is 39,900 watts.

Maximum power occurs at force times maximum velocity. So maximum power is 79,800 Watts.

 

[lukeodn]

Im pretty sure that the dagree of incline is one tenth of a dagree, because it says 1 in 10.

So has the tractive force actually been calculated?

Is the diagram being drawn to calculate the normal force? if not i am confused at to it's purpose?

Thankyou for you help, much appreciated

 

[RTW69]

If the slope is 1:10 the slope angle would be 5.7 degrees because the slope rises 1 unit for every 10 units of run therefore Tan (theta)= 1/10

The free body diagram identifies all the forces on the truck and breaks them into components parallel and perpendicular to the slope. The forces acting parallel to the slope must equal must equal mass*acceleration acting up the slope or 3000 Newtons. The forces acting on the truck parallel to the slope are the forces due to gravity acting in a negative direction, the friction force also acting in a negative direction and the force required to accelerate the truck up the slope (positive direction).

To determine the friction force you need the normal force which is a force acting perpendicular to the slope. The free-body diagrams help visualize the geometry of the forces. In this case the normal force will be m*g*cos(5.7 degrees) or 14,627 Newtons. The friction force therefore equals the normal force times u_k or 731 Newtons (by definition). The force due to gravity acting down the slope is 1460 Newtons. Do you see how to get that value?

 

[lukeodn]

Oh, i see what your saying now about the diagram.

What are you saying that uk is?

And no i don't really see how you get that value? I am getting the rest though!

Thanks again.

 

[RTW69]

In your problem you state the coefficient of friction is 0.05. That value is the coefficient of kinetic friction or u_k. the friction force is u_k*the normal force.

The truck is on a slope. The force due to the mass of the truck is acting toward the center of the earth. That force must be resolved into forces that are perpendicular and parallel to the slope. We must us a little trigonometry to do so. That is where the sin(theta) and cos(theta) values come from. Just about any physics book will show you how to draw free-body diagrams of blocks moving on sloped surfaces.

 

[lukeodn]

yes i understand now!

Thankyou mate, BIG HELP!