Rotational motion of rear wheels

[TeeNaa]

A 1500 kg car is driven onto a pair of ramps so that it fronts wheels have been raised 0.3 m above the rear wheels. The wheel base of the car is 2.4m and it center of gravity is located at the midpoint between the front and rear wheels. How much of the car's weight is supported by the rear wheels? - The problem I'm having in this question is interpreting what the question is asking for. I just need on on setting up the problem as on what I should find first. Thanks

This is what I drew but I have no idea what to use to solve with these given information.

 

[Simon Bridge]

When the car is level, it's weight is distributed evenly between the front and rear wheels, but when the front is lifted higher, more weight gets pushed onto the back than the front. Can you see why?

What has happened to the weight vector, compared with the wheel positions, to make that happen?

If you draw a free-body diagram of the situation - you get two normal forces: one from the front wheels and one from the back wheels. You also get horizontal surface forces at each wheel-set.

 

[TeeNaa]

@Simon, thank you for the reply.
For the first question, yes I can see why the weight would be pushed back. For the second question, I don't know what that really mean. I understand why there is two normal force because of the wheel set but I do not understand the horizontal surface forces part. What would I have to find first in this case? Thanks again.

 

[Simon Bridge]

The question involves a free body diagram where not all the forces act through the center of mass.
Other than that, it is still statics.

 

[Nickie]

Based on the information provided, it seems like the question is asking for the amount of weight that is being supported by the rear wheels of the car. To solve this problem, you will need to use the concept of rotational motion and equilibrium. Here are the steps you can follow to solve this problem:

1. Draw a free body diagram of the car, showing all the forces acting on it. This will include the weight of the car (mg), the normal forces from the front and rear wheels, and the forces from the ramps.

2. Since the car is in rotational equilibrium, the sum of all the torques acting on it must be equal to zero. The torque due to the weight of the car will depend on the distance between the center of gravity and the point where the weight is acting. In this case, since the center of gravity is located at the midpoint between the front and rear wheels, the distance from the center of gravity to the weight is 0.3m/2 = 0.15m.

3. The torque due to the normal forces from the front and rear wheels will depend on their respective distances from the center of gravity. Since the wheel base is 2.4m, the distance from the center of gravity to the front wheels is 2.4m/2 = 1.2m. Similarly, the distance from the center of gravity to the rear wheels is also 1.2m.

4. Now, set up the equation for rotational equilibrium:
Sum of torques = 0
(mg)(0.15m) + (normal force from front wheels)(1.2m) + (normal force from rear wheels)(1.2m) = 0

5. We know that the weight of the car is equal to its mass (1500kg) multiplied by the acceleration due to gravity (9.8m/s^2). So we can substitute this into the equation:
(1500kg)(9.8m/s^2)(0.15m) + (normal force from front wheels)(1.2m) + (normal force from rear wheels)(1.2m) = 0

6. Since the car is in equilibrium, the sum of the normal forces from the front and rear wheels must be equal to the weight of the car. So we can rewrite the equation as:
(1500kg)(9.8m/s^2)(0