Calculating Normal Reaction & Contact Force on 800kg Van Attached to Car

AI Thread Summary
The discussion focuses on calculating the normal reaction and contact forces acting on an 800 kg camping van attached to a car during acceleration. Participants clarify that the car accelerates at approximately 4.17 m/s² to reach 60 km/h in 4 seconds, leading to a force calculation using Newton's second law. The normal reaction force acts vertically under the wheels, while the contact force at point P has both horizontal and vertical components. Suggestions include taking moments about point P for easier torque calculations, confirming that this approach simplifies the analysis. The conversation emphasizes understanding the forces and their directions to solve the problem effectively.
bluewood
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Homework Statement


The following camping van with a total mass of 800 kg is connected at point P to the back of a car. The car accelerates with constant acceleration a during 4 seconds from rest until it achieves 60 km/h. Assuming there is no resistance (from air and ground), calculate the normal reaction in the wheels and the components of the contact force acting at point P.

Homework Equations





The Attempt at a Solution



I'm really confused in this one, so any help to start would be nice.
 

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bluewood said:

The Attempt at a Solution


I'm really confused in this one, so any help to start would be nice.
Let's start simple then.


bluewood said:

Homework Statement


The following camping van with a total mass of 800 kg is connected at point P to the back of a car. The car accelerates with constant acceleration a during 4 seconds from rest until it achieves 60 km/h.

So by Newton's 2nd law, how much force is the car producing and in what direction is it (use x and y for the horizontal and vertical directions)

bluewood said:
Assuming there is no resistance (from air and ground), calculate the normal reaction in the wheels and the components of the contact force acting at point P.

On the wheel, what direction is the normal reaction?

At the contact force, there will be a reaction in two directions, what directions will those be ? (vertical and/or horizontal?)
 
The force in the x direction that makes the van move:

v = \frac{60 \times 10^3}{3600} m/s \approx 16.667 m/s

a = \frac{16.667}{4} = 4.16675 m/s^2

F_x = 800 \times a

I suppose the normal reaction acts in the y direction and is located under the wheels. The F force acting at point P is made of two components Fx and Fy acting in x direction and y direction, respectively. Fx is the force previously calculated.

The motion equation in the x direction is:
F_x = m a (already solved)

and in the y direction:
F_y + R_n - W = 0

Now the torque equation is missing. But I don't know where to put the axis of rotation and calculate the corresponding torques.
 
bluewood said:
Now the torque equation is missing. But I don't know where to put the axis of rotation and calculate the corresponding torques.

You can put it anywhere you want, remember if it is in equilibrium, the sum of moments about any point is zero.

So you can take moments about the contact point , or the normal reaction of the wheels or the weight. Though I'd suggest about the contact point P.
 
Something like this?

1.2 \times W - 1.2 \times R_n = 0

1.2 \times 800 \times 9.8 - 1.2 \times R_n = 0
 
Yes that should work out correctly.
 
You have three forces on the van, W FP and FC.

The question asks for FP, but not for the whole of FC. so it's quickest to take moments about C

that immediately gives you the direction of FP, doesn't it? :wink:
 
tiny-tim said:
You have three forces on the van, W FP and FC.

The question asks for FP, but not for the whole of FC. so it's quickest to take moments about C

that immediately gives you the direction of FP, doesn't it? :wink:

Wow!Take the axis at point P is very easy to calculate. Thanks .
 
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