Basic Apparent Weight Question

In summary, a person weighing 60 kg rides in a 20 kg cart moving at 13 m/s at the bottom of a valley with a radius of 36 m. The apparent weight of the person is calculated to be -306.33 N, indicating that the person feels lighter due to the opposing forces of gravity and centripetal force. The direction of these forces depends on the chosen coordinate system. The inclusion of the mass of the cart does not affect the calculation of the apparent weight.
  • #1
RJLiberator
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Homework Statement


A 60 kg person rides in a 20 kg cart
moving at 13 m/s at the bottom of a valley that is in the shape of an arc of a circle
with a radius of 36 m. What is the apparent weight of the person as the cart passes
the bottom of the valley?

Homework Equations


F_centripital = m*v^2/r
F = ma

The Attempt at a Solution



So this seems a little too easy to be correct.
Apparent weight = m(a+g)=m(v^2/r+g)
=60(13^2/36+9.80)
=869.67 N

The thing I am wondering about is the sign on gravity according to it being in the bottom of a valley.
Is my work correct?
 
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  • #2
Question, where is the centripetal force directed and where is the gravitational force directed?
 
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  • #3
The centripital force is directed upwards and the gravitation force is directed downwards.

This would imply that the correct equation would be: m(a-g) = m(v^2/r-g) ?
 
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  • #4
Yes! or (g-a) depending on how you chose your coordinate system.
From this, we will have a smaller magnitude of apparent weight. This creates the feeling of lightness when you are going fast down a hill in your car, or a situation similar.
 
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  • #5
60(13^2/36-9.80)
nets me an answer of -306.33N
The negative apparent weight implies that the body is being accelerated downward with an acceleration greater than that exerted by the Earth.
Is this a correct understanding?
 
  • #6
Well, if you considered just the weight of the person stationary, 60 x -9.8 = -588 N. You computed -306.33 N, he feels less of a pull then he would normally by gravity, i.e. feeling lighter.

The forces oppose each other so their sum cannot have a magnitude greater than the individual forces.
 
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  • #7
Ahhhh, now that makes sense to me.

So regularly, he is being pulled at -588 N. In my original answer of 806 he would have been feeling heavier which wouldn't make sense.
Since we updated it to be -306.33 N, we see that he is feeling relatively lighter.

Great. Thank you kindly for your help.
 
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  • #8
RJLiberator said:
Ahhhh, now that makes sense to me.

So regularly, he is being pulled at -588 N. In my original answer of 806 he would have been feeling heavier which wouldn't make sense.
Since we updated it to be -306.33 N, we see that he is feeling relatively lighter.

Great. Thank you kindly for your help.

No problem, happy to help!
 
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  • #9
Two more things regarding this thread:
1) Why was the 20kg of the cart neglected throughout this entire problem? Was this thrown in just to throw me off?
2) The force vectors in a full-body diagram would be: Force of gravity pointing downwards which would be equal to the centripetal force which points upwards towards the center of the "circle" and then a velocity vector pointing to the right perpendicular and a friction vector pointing the opposite way, but smaller than the velocity vector.
Correct?
 
  • #10
RJLiberator said:
A 60 kg person rides in a 20 kg cart
moving at 13 m/s at the bottom of a valley that is in the shape of an arc of a circle
with a radius of 36 m. What is the apparent weight of the person as the cart passes
the bottom of the valley?
1) Looking at the problem again, we should have done 60 + 20 for the mass the entire time. So readjust your calculations with 60+20 as the mass. I went off the number you used originally (60) in your attempted answer so I did not notice that there was the 20 kg cart.
EDIT: nevermind, it asks for the apparent weight of the person and not the apparent weight of the cart + person system.

2) If you are just drawing the force vectors, do not include the velocity vector since that is not a force. As with friction, that is a force so it would be appropriate to draw that IF friction is applied to the problem (which it doesn't specifically mention).
 
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  • #11
I'm looking over this problem again, trying to understand everything:

A) So the stationary person feels a -784N force? How can someone feel a negative force?

And then B) When he is at the bottom of the cliff we calculated a -408.4N force and since this is more in the positive direction, this is a heavier force?

Hm
 
  • #12
Negative meaning opposite the direction you chose to be positive (ex up is positive so down is negative, or down is positive so up is negative)

For the second paft, the magnitude of the resulting force is less (400 < 700) so he is less pulled down then he would normally be.
 
  • #13
Ah, the absolute value will dictate the force. The orientation of my coordinate system dictates whether it will be negative or positive. Thank you kindly.
 
  • #14
No problem, happy to clarify!
 

What is basic apparent weight?

Basic apparent weight is the perceived weight of an object when it is placed on a scale. It takes into account the effect of the object's mass and the force of gravity acting on it.

How is basic apparent weight different from actual weight?

Actual weight is the true measurement of the gravitational force acting on an object, while basic apparent weight is the perceived weight of the object on a scale. Basic apparent weight takes into account any other forces acting on the object, such as buoyancy or acceleration.

What factors affect basic apparent weight?

The main factors that affect basic apparent weight are the object's mass and the strength of the gravitational force acting on it. Other factors that can affect it include the density of the surrounding medium and any other forces acting on the object, such as air resistance or acceleration.

How is basic apparent weight measured?

Basic apparent weight is typically measured using a scale or balance. The object is placed on the scale and the reading is taken, which represents the basic apparent weight of the object.

Why is understanding basic apparent weight important?

Understanding basic apparent weight is important in many fields of science, such as physics, engineering, and chemistry. It allows us to accurately measure and predict the behavior of objects in different environments, and is essential in designing structures and machines that can withstand various forces and conditions.

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