Magnitude of the average resultant force in KN on the truck

[remaan]

Homework Statement

A 2000 kg truck traveling at a speed of 6 m/s makes a 90 turn in a time of 4 s and emerges from this turn with a speed of 4 m/s. What is the magnitude of the average resultant force in KN on the truck during this turn?

Homework Equations

F avr. = Pf - Pi / t

The Attempt at a Solution

Pf = m Vf = 2000 * 4 = 8000

Pi = m Vi = 2000 * 6 = 12000

Pf - Pi = - 4000 and F avr. in KN = -1

Is that Right ?

 

[remaan]

I was wondering if it's correct to use the momentom equation in such a case, where we have only one object ?

And how can the"90 degree" effect the solution ?

 

[djeitnstine]

Yes that's correct. 90 degrees does not affect anything, it could have been 180 degrees and the resulting average force would have been the same.

 

[remaan]

What do you think of what I did ?

 

[djeitnstine]

It is perfectly fine. Another way to think of it is finding the average deceleration (vf-vi/t) then multiplying it by the mass of the truck. However, look closely and it is essentially the same thing you did by conservation of momentum.

 

[remaan]

Uha, this gives the same answer, 1000, which is not in the choises,

Or 1 Kn
and that

what cofuses me.

 

[Dopefish1337]

What are you getting as a result for ((v_f-v_i)/t)?

edit: Your answer does seem to be correct, I just mispushed a button when I did it. Odd.

edit2: Actually, consider the direction of each of the momentums. Think pythagoras.

 

[remaan]

Do you mean to find the velocity components ?

I tried that, but with no difference because the angle is 90 ..

What do you think?

 

[Dopefish1337]

Momentum is a vector itself (it has direction), so you don't need to break it down to just velocity. The 90 degrees bit makes it relatively easy to compute the momentum needed. Try drawing a triangle showing the initial and final momentums.

 

[remaan]

Yes, and its direction depends on the Vel. direction ??


Besides, how can I draw the triangle showing the momentom ?
its a quarter from a circle ?
Right ?

 

[Dopefish1337]

Yes, the direction comes from velocity. For simplicity, it may be easiest to assume the initial momentum is pointing north, and after turning, it's going east (although any corodinate system where they're 90 dergees apart will do).

If you place the two momentum vectors tail to tail at a common origin, what is the size of the vector connecting the two tips?

 

[remaan]

Ok, very good, I did that and I got

a vector R which magnitude is vf - vi

But when I multlied its magnitude by the mass and divide by 4 which is the time to find F avr.

I got a very big number ?

 

[Dopefish1337]

How did you conclude that its magnitude was vf-vi?

 

[remaan]

Or,

I was thinking about that way :
deltap = pf - Pi
mVf- mVi
m(Vf+Vi) assuming that this direction is positive

then subs.

2000( 6+4) = 20,000/4 = 5000 N

What do you think ??

 

[Dopefish1337]

deltap = pf - Pi

This is only true if the vectors were in the same direction. In this case, they're perpindicular to one another.

mVf- mVi
m(Vf+Vi) assuming that this direction is positive

If you factor out an m, you would have m(vf-vi), not m(vf+vi), although from before, I don't believe you can just subtract them as you're doing.

 

[remaan]

Then, just tell me how to the triangles ??

I mean after I draw the triangle what shall I do next ??

 

[Dopefish1337]

One of your vectors will point up like "|", and the other sideways like "_", so you have the base and height of a right angled triangle. You want the hypotenuse.

 

[remaan]

Then, what does it mean?

Is it the momentom ?

Or it's vf- vi

 

[Dopefish1337]

If the vectors you use are momentum vectors, than the hypotenuse will be momentum as well. (It would be possible to do it with velocities as well, but that's more work.)

 

[remaan]

Well, what I did is I multplied this hyp. by the mass and divide by the time

Right ?

 

[Dopefish1337]

If you used velocities instead of momentums for your base and height, then yes.

If you used momentums, there's no need to multiply by mass.

 

[remaan]

Uha, Thanks, I got that ! :D

 

[danb]

magnitude of the average force

The first thing you have to deal with is that the magnitude of the average force isn't necessarily the same thing as the average magnitude of the force over time. To calculate the average of the magnitude, you would have to know the details of how the force varies over time.

The magnitude of the average is easier. Just find the average vector force and calculate its magnitude.

Defining the x-axis as the truck's initial direction and the y-axis as its final direction, the average vector force on the truck is
F_avg = delta p / delta t
and
delta p = m delta v = m (v_f - v_i) = m(v_fj - v_ii)
So the average vector force is
F_avg = (m / delta t) (v_fj - v_ii)
and its magnitude is
F_avg = (m / delta t) sqrt(v_f² + v_i²)

 

[remaan]

So do you think that the answer can be 3.6 K N ??