Normal Force acting on tries of a go-cart

In summary, the total weight of the go-cart and driver is 1068N and their combined center of mass is shown in the attachment. The rear wheels exert a 106.7 N horizontal force together on the track, while the horizontal force on the front wheels is neglected. Using the equation NA+NB-W=0 and the moment around CM in the clockwise direction (NA*(0.4064)+F*(0.381)-NB*(1.1176)=0), the normal forces exerted on the tires at A and B were solved to be NA = 788.29N and NB = 279.7N, respectively. However, after considering the direction of the torque from each force, the correct answer was
  • #1
Hanga
14
0

Homework Statement


The total weight of the go-cart and driver is 1068N. The location of their combined center of mass is shown in attachment. The rear wheen exert a 106.7 N horizontal force together on the track. Neglect horizontal force on the front wheels.

a) What is go carts acceleration? [SOLVED MYSELF]
b) What are the normal forces exerted on the tires at A and B?

Homework Equations


Moment around centre of mass.


The Attempt at a Solution


NA+NB-W = 0 (where NA and NB is normal force acting on the tires)

Moment around CM in clockwise direction
NA*(0.4064)+F*(0.381)-NB*(1,1176) = 0

solving for NA and NB I get
My answer; NA = 756,52 N NB = 485 N, which is wrong according to the book.

I don't know what is wrong. Please help.
 

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  • #2
Hanga said:
Moment around CM in clockwise direction
NA*(0.4064)+F*(0.381)-NB*(1,1176) = 0
What orientation (sign) is the torque from each force?
 
  • #3
Doc Al said:
What orientation (sign) is the torque from each force?

Not quite sure what you are getting at, but I can take either clockwise or counter clockwise as direction goes. Even if I take it counterclockwise it turns out wrong..
 
  • #4
Hanga said:
Not quite sure what you are getting at, but I can take either clockwise or counter clockwise as direction goes. Even if I take it counterclockwise it turns out wrong..
There are three forces acting. Describe the direction of the torque from each, clockwise or counterclockwise. (You're messing up your signs in your equation.)
 
  • #5
Ok so I checked so I believe it should be F+NA-NB as the force is below the centre of mass the cart want to rotate backwards. Doing this I get NA = 788,29 NB= 279,7, this is fairly close to the books answer but it's not correct.
I'm not hunting for an answer tho, just the correct way of thinking. I thought I understood this type of question, but I seem to get it wrong every time :P
Please continue to guide me through this! :)
 
  • #6
Hanga said:
Ok so I checked so I believe it should be F+NA-NB as the force is below the centre of mass the cart want to rotate backwards.
How is this different from what you had before?
Please continue to guide me through this! :)
I'm trying to! :smile: Again: Describe the direction of the torque from each force as clockwise or counterclockwise.
 
  • #7
I'm not sure what you mean, I am afraid. I have no resource to study this, but i'll try.
You want me to describe direction of torque from each force as cw, or ccw.

NA would be CCW, as I stated its because it's below centre of mass.
NB would be CCW.
The force F is probably what I am having problem with, and it's direction. The force could possibly be CCW aswell. I have an exact answer from a home page I found and it's attached, there they state the force of torque around CM as described in attachment. Its so weird as I am only +20 N away from the correct answer, and no matter how I choose to change the forces from CCW to CW it won't change just 20N.
I did the exact problem with a refrigirator and the only general difference is that a) It does not have wheels, 2) NA and NB is on the edges of the refrigiratior on the floor. The only thing not used in this problem is the radii of the wheels, but how can they affect the problem? I am getting a headace...
 

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  • #8
Hanga said:
NA would be CCW, as I stated its because it's below centre of mass.
Incorrect. You must consider the direction of the force, not just where it acts. If Na were the only force acting, which way would the cart tend to rotate?
NB would be CCW.
Good.
The force F is probably what I am having problem with, and it's direction. The force could possibly be CCW aswell.
No 'possibly' about it--definitely. Again, if F were the only force acting, which way would the cart tend to rotate?
 
  • #9
Doc Al said:
Incorrect. You must consider the direction of the force, not just where it acts. If Na were the only force acting, which way would the cart tend to rotate?

NA should def be CW as it acts upwards from the floor.

Doc Al said:
No 'possibly' about it--definitely. Again, if F were the only force acting, which way would the cart tend to rotate?
The cart must rotate CCW. Just as a fast moving car or motorcycle which accelerate fast enough would rotate CCW. So my final answer is that NA is CW and F and NB is CCW in this case.

What would be the appropriate next step?
 
  • #10
Hanga said:
NA should def be CW as it acts upwards from the floor.
Good.
The cart must rotate CCW. Just as a fast moving car or motorcycle which accelerate fast enough would rotate CCW. So my final answer is that NA is CW and F and NB is CCW in this case.
Good.
What would be the appropriate next step?
Set up a net torque = 0 equation just like you did before. (Assign one orientation as positive, the other as negative.)
 

Related to Normal Force acting on tries of a go-cart

What is normal force acting on the tires of a go-cart?

The normal force acting on the tires of a go-cart is the force that the ground exerts on the tires in a direction perpendicular to the surface of the ground. This force is responsible for keeping the go-cart from sinking into the ground and for providing the necessary friction for the tires to move the go-cart forward.

What factors affect the magnitude of normal force on the tires of a go-cart?

The magnitude of normal force on the tires of a go-cart is affected by the weight of the go-cart, the weight distribution of the go-cart, the type of tires used, and the surface on which the go-cart is being driven. As the weight of the go-cart increases, so does the normal force. Additionally, a larger weight distribution on the rear tires results in a larger normal force on the rear tires.

Does the normal force change during a turn?

Yes, the normal force changes during a turn. When a go-cart is turning, the weight distribution shifts towards the outside of the turn, resulting in a larger normal force on the outside tires and a smaller normal force on the inside tires. This is why go-carts tend to lean towards the outside of a turn.

How does the normal force affect the handling of a go-cart?

The normal force plays a crucial role in the handling of a go-cart. It determines the amount of friction between the tires and the ground, which affects the go-cart's ability to accelerate, decelerate, and turn. A larger normal force allows for better traction and control of the go-cart.

Can the normal force be greater than the weight of the go-cart?

Yes, the normal force can be greater than the weight of the go-cart. This can happen when the go-cart is accelerating or going up a hill, as the tires need to exert a greater force on the ground to overcome the forces acting against the go-cart. However, the normal force can never be less than the weight of the go-cart, as the go-cart would sink into the ground without enough normal force to support it.

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