What direction will the force from acceleration act in this scenario?

[t_n_p]

Homework Statement

I have a 60kg cart on the back of my truck (travelling at 80km/h) which falls off the back and slides 45m before coming to a stop. I want to find coefficient of kinetic friction

http://img9.imageshack.us/img9/4356/38603897.jpg

Homework Equations

Vf = 0km/h
Vi = 80km/h = 200/9 m/s
s = 45m

using Vf^2 = Vi^2 + 2as,
I find acceleration to be = - 4000/729 m/s^2 which I think is correct up to this point as the negative indicates decelleration

This is where I get confused,
I draw my free body diagram, which way will the force created by acceleration (i.e. F=ma) force act? Will it be to the right or to the left?

 

[LowlyPion]

Vf = 0km/h
Vi = 80km/h = 200/9 m/s
s = 45m

using Vf^2 = Vi^2 + 2as,
I find acceleration to be = - 4000/729 m/s^2 which I think is correct up to this point as the negative indicates decelleration

This is where I get confused,
I draw my free body diagram, which way will the force created by acceleration (i.e. F=ma) force act? Will it be to the right or to the left?

Your velocity is left. The acceleration is right. The friction acts against the direction of motion.

 

[t_n_p]

so if i draw a FBD...

http://img10.imageshack.us/img10/9336/41243042.jpg

frictional force = µN
I know N, I want µ, which means I need to know frictional force.
is frictional force equal and opposite to the F=ma force?

my qualm is that if it is equal and opposite, doesn't this suggest the box is static, rather than moving?

 

[DClancy]

Once the box falls off the vehicle the force of friction is the force responsible for the acceleration. So in your case once it hits the ground µN=ma and it will be static once it comes to rest because the deceleration from friction is a reactionary force that resists motion.

 

[LowlyPion]

I know N, I want µ, which means I need to know frictional force.
is frictional force equal and opposite to the F=ma force?

my qualm is that if it is equal and opposite, doesn't this suggest the box is static, rather than moving?

No as DClancy says what you have labeled m*a is your µ*m*g. Your situation is not static if there is acceleration. It's only static when it comes to rest 45m later. So your a = µ*g and it is directed to the right in your drawing.

 

[t_n_p]

right, so its more like...

http://img15.imageshack.us/img15/9466/83450616.jpg

which results in µ=a/g

I'm just a little confused as to how velocity can be left and acceleration can be to the right
thanks for the help

 

[DClancy]

Ok, your diagram looks good. So now let's think about what is happening. The box falls off the truck and right before it hits the ground and begins to slow it is going 80 km/h. That velocity is in the same direction the truck was going, based on the orientation of your diagram the velocity is to the left. Once the box touches the ground the force of friction acts on it, the force of friction is directed to the right, opposite of the velocity, since the box comes to a stop the FF must be opposite the velocity in order to slow it down and eventually stop it.

 

[t_n_p]

ok cool, that makes things clear.

thanks for the help guys

 

[Count Iblis]

There is a huge mistake in the problem. The answer depends also on the height from which the cart falls. Suppose that the cart falls from a height of H. Then the velocity in the vertical direction just before the cart hits the ground, Vv, satisfies by conservation of energy:

1/2 Vv^2 = g H -------->

Vv = sqrt[2 g H]

When the cart makes contact with the ground, there will be a very large normal force acting on the cart for a short time (tau) which will cause the momentum of the cart in the vertical direction to become zero. tau will be of the order of the thickness of the cart divided by the speed of sound. The integral of the magnitude of the normal force over this short time will be equal to

m Vf +m g tau= m sqrt[2 g H] + m g tau

Since tau will be very small, we can ignore the last term. Now, mu times the normal force is the friction force. So, the integral of the friction force over the time from the moment when the cart makes contact with the ground till a time tau later when the vertical motion of the cart has relaxed to zero is given by:

mu m sqrt[2 g H]

This is equal to the change of momentum in the horizontal direction. So, the speed decreases by an amount:

mu sqrt[2 g H]

This then means that Vi should be replaced by:

Vi - mu sqrt[2 g H]

This then gives you a quadratic equation for mu.

 

[t_n_p]

no value for height is given in the question, therefore for all intensive purposes I think it has been ignored

 

[Count Iblis]

no value for height is given in the question, therefore for all intensive purposes I think it has been ignored

Yes, but if you take H to be of the order of 1 meter, you see that it already has a very significant effect. Thing is that the cart is dropped and has to stop moving in the vertical direction, so it is an essential effect and not at all like ignoring air friction in ball throwing problems.

A similar mistake in a similar problem was spotted by me here:

https://www.physicsforums.com/showthread.php?t=297204

 

[t_n_p]

well i have no value to work with, what would you do?

I will seek clarification in the meantime

 

[Count Iblis]

well i have no value to work with, what would you do?

I will seek clarification in the meantime

I would first explain in the solution that the height is important. Then I would simply assume a realistic value for the height , like H = 2 meters. Then I work out the solution.

What matters is that you understand the physics of the problem and are able to compute the answer. In some cases a teacher can overlook an essential part of the physics involved. So, if a student spots that and changes the formulaton of the problem to take that into account, then all the teacher can do is give that student some bonus points.

 

[LowlyPion]

So, if a student spots that and changes the formulaton of the problem to take that into account, then all the teacher can do is give that student some bonus points.

There is also the counter view that a teacher might do some off balance sheet accounting at grade time and evaluations if a student has raised specious and smart alec considerations and disrupts the learning process for others.

One would be advised to balance the risks I'd think.