Inertia and moving plane problem

In summary, the homework statement states that a moving vehicle of mass M is moving down an inclined plane of angle alpha with respect to the horizontal plane. There is a force indicated on the scale which is written as pèse-personne between M and m.
  • #1
gillouche
Gold Member
25
7
Hi,

I have some troubles to understand the next step for the solution. If you need more informations, please let me know.

1. Homework Statement


A moving vehicle of mass M is moving down an inclined plane of angle alpha with respect to the horizontal plane. What is the force indicated on the scale ?

NcEEozh.png


It is written pèse-personne between M and m. That's a scale (to weight people)

Homework Equations


[/B]
Acceleration of M on the inclined plane : a = g*sin(alpha) * (M/m + 1)

The Attempt at a Solution


[/B]
I started to try to find the acceleration of M on the inclined plane. I have the following forces acting on my moving vehicle : gravity (down), normal force (perpendicular to the inclined plane) and the weight of the mass m (it is not indicated but I assume that the scale doesn't have a mass).

I separated the gravity components in parallel and perpendicular component and I get :

Fg parallel inclined plane = (M+m)g * sin(alpha)
The Fg perpendicular cancels out with normal force. I don't think I need it but that's the same of Fg parllel but with cosine.

Then I have ma = (M+m)g * sin(alpha) which gives me a = g*sin(alpha) * (M/m + 1)

(I can give the steps if needed).

Now I am stuck. How can I find the normal force for m ?

For m, I have the weight (down), the normal force (up), acceleration (right) and inertia (left). Those orientations are with respect to the moving vehicle and not the inclined plane. Am I supposed to break all the components in their parallel and perpendicular (to the inclined plane) components ?

I am confused with this inclined and not inclined plane of the moving vehicle. The exercises we did before was an elevator (for the weight) and a moving vehicle on the horizontal axis (to introduce inertia).

Thanks.
 
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  • #2
Perhaps take a fresh look at this.

Do you think horizontal motion will affect the reading on the scales?
 
  • #3
No, only the gravitational force acting on m + the vertical acceleration that m has (other than g).

So I have my acceleration to the right parallel to the inclined plane, I should find the total acceleration perpendicular to the horizontal plane for m.
 
  • #4
If horizontal motion is not relevant, then how can you simplify the problem?

Hint: Acceleration is a vector.
 
  • #5
I know that my acceleration vector for the moving vehicle is parallel to the inclined plane. I can use the angle alpha to break it into x and y components. For the y component (that's the one I need for the vertical acceleration), I get g * sin2(alpha) * (M/m + 1)

If I look at the forces acting on my mass m. Is that correct to say that I have the following :

ma = mg + m * (perpendicular acceleration component of a) - normal force for m

So I should replace the "a" by the acceleration of M on the inclined plane and "perpendicular acceleration component of a" then simplify ?

I did it and I get mg(1 - sin(alpha) * (M/m + 1) + sin2 * (M/m + 1)))

The answer I get seems weird.
 
  • #6
I can't understand why you think M is relevant. If you are standing on scales in your house, is the mass of your house relevant?
 
  • #7
PeroK said:
I can't understand why you think M is relevant. If you are standing on scales in your house, is the mass of your house relevant?
No but the acceleration of the whole system depends on M too. Am I wrong ? Maybe my calculation for the acceleration of the whole system is wrong.
 
  • #8
gillouche said:
No but the acceleration of the whole system depends on M too. Am I wrong ? Maybe my calculation for the acceleration of the whole system is wrong.
The acceleration of the whole system depends neither on M nor m. Only on g and the angle of inclination.

And the force m exerts on the scales cannot depend on the mass underneath the scales.
 

1. What is inertia and how does it relate to the moving plane problem?

Inertia is the tendency of an object to resist changes in its state of motion. In the context of the moving plane problem, it refers to the resistance of an object to changes in its velocity and direction of motion.

2. How does Newton's first law of motion apply to the moving plane problem?

Newton's first law of motion, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will remain in motion with a constant velocity unless acted upon by an external force. This directly applies to the moving plane problem as the objects on the plane will continue to move in a straight line at a constant speed unless a force, such as air resistance, acts upon them.

3. What factors can affect the inertia of objects on a moving plane?

The inertia of objects on a moving plane can be affected by the mass and shape of the objects, as well as the speed and direction of the plane. Objects with larger mass and more streamlined shapes will have greater inertia and be more resistant to changes in motion.

4. How does the concept of centripetal force relate to the moving plane problem?

Centripetal force is a force that acts towards the center of a circular path, causing an object to move in a curved path. In the context of the moving plane problem, the centripetal force is responsible for keeping objects on the plane moving in a circular path, such as during a turn or loop-the-loop.

5. What is the difference between inertia and momentum in the moving plane problem?

Inertia refers to an object's resistance to changes in its state of motion, while momentum refers to an object's quantity of motion, taking into account both its mass and velocity. In the moving plane problem, inertia affects an object's ability to change its state of motion, while momentum determines the strength and direction of that change.

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