1. ## Mechanics

I'm sitting a Mechanics paper as one of my Maths modules tomorrow and I'm stuck on a practice question, please can someone help if they can.

A car is towing a trailer along a horizontal straight road using a horizontal tow-bar. The masses of the car and the trailer are 1050kg and 200kg respectively. The resistance to motion of the car is 850N and the resistance to motion of the trailer is 150N.

Question
At an instant when the driving force exerted by the car is 1100Nm find,
a) the acceleration of the car
b) the pulling force exerted on the trailer

At another instant the pulling force exerted on the trailer is zero
a) show that the acceleration of the car is -0.75m/s/s
b) find the driving force exerted by the car

a) F = ma
1100 = (1050 + 200)a
1100/1250 = a = 0.88m/s/s

Do I need to use the resistances in the above?

b) What is the pulling force? Is it like a resistive force?

Um, and I don't understand the rest
Please can someone explain it to me.

Many thanks,
Delta

2. ## Sum of the forces = ma

Actually you should think of the equation as:

$\displaystyle \Sigma F = ma$

So yes you need to take account the resistance

$\displaystyle \Sigma F = 1100Nm - 850Nm - 150Nm = 100Nm$

$\displaystyle 100Nm = 1250Kg*a$

$\displaystyle a = 0.08 m/s^2$

3. ## Free Body Diagram FBD

b) if you do a free body diagram on the trailer then it has three forces on it:

1. its weight pointing straight down.
2. The pulling force exerted on it by the truck
3. The resistive force in the opposite direction of the pulling force.

The point of this question is for you to realize that the force on the trailer is less then the force exerted by the truck because the force on the truck is being spent to overcome the resistive force on the truck and to accelerate the truck.

4. ## F = ma

For the second question you need to show that the resistive forces are deccelerating the vehicle at the given rate using $\displaystyle \Sigma F = ma$

5. Thank you very much. I thought it was a bit pointless putting in those resistive forces if I hadn't used it!

Please can you tell me what the pulling force is in the question. Is it a force in the opposite direction to the driving force?

Thanks,
Delta

6. ## pulling force

Imagine a rope between the truck and the trailer.

When the truck pulls on the rope, the pulling force is exerted on the trailer. The pulling force the trailer feels is the force you would replace the truck with in a free body diagram.

I can always explain further if you need more help.

7. I'm not sure if I understand properly. So in part b) of the first question, would I need to use the weight as well to find the answer? I'm really confused.

Thanks,
Delta

8. no ignore the weight. it would be proper for lableing the FBD but not needed for this problem.

You just need to take the force exerted by the car minus the resistive force of the car - mass*acceleration of the car = the pulling force applied to the trailer.

9. Okay, thank you.

So in the second question, I would need to tackle it in a similar way - just make the pulling force zero?

Many thanks,
Delta

10. ## pulling force

The resistive forces are the only ones acting on the car and trailer.

There should be no pulling force on the trailer but do not confuse the two issues. The pulling force question was just a detail of the previous problem. Now we want to talk about the car and trailer together as 1 system again so we do not really need to think about the pulling force.

11. I can't figure it out. I think I'm going to have to leave it and ask when I get to school. Just a few more hours until my exam

Thank you for all your help.
Delta

12. Here is one way.

"resistance to motion" here is inertia.
When the object is not moving, the resistance to motion is inertia at rest. It's direction is opposite to that of the force that tries to move it.

When the object is moving, the resistance to motion is inertia in motion. It's direction is the same as that of the force moving it.

In your question, the car and trailer are already moving, so the resistances to motion of the two are in the direction of where the car and trailer are going.

Note: since the two are attached to each other, then the acceleration of the trailer is always the same as the acceleration of the car.

-----------
Question
At an instant when the driving force exerted by the car is 1100Nm find:

>>>a) the acceleration of the car.
F = m*a
where
F = the driving force plus the resistances to motion = 1100 +850 +150 = 2100 N
m = total mass of the system = 1050 +200 = 1250 kg
So,
2100 = 1250*a
a = 2100/1250 = 1.68 m/sec/sec. ----answer.

>>>b) the pulling force exerted on the trailer.
Let P = pulling force exerted on the trailer, in N.
F = m*a
where
F = P +(resistance to motion of trailer) = P+150
m = mass of trailer = 200 kg
a = 1.68 m/sec/sec
So,
P+150 = 200*1.68 = 336
P = 336 -150 = 186 N ----answer.

----------
At another instant the pulling force exerted on the trailer is zero.
Meaning, P = 0.

>>>a) show that the acceleration of the car is -0.75m/s/s.
See the part (b) above.
F = m*a
where
F = P +(resistance to motion of trailer) = 0+150 = 150 N
m = mass of trailer = 200 kg
a = ?
So,
150 = 200*a
a = 150/200 = 0.75 m/sec/sec

Meaning, the car and the trailer are accelerating at 0.75m/sec/sec when the car is not exerting any pull on the trailer.
Why then is the aceleration negative? -0.75m/sec/sec?

>>>b) find the driving force exerted by the car.

The system is in motion, the car is not pulling the trailer, the acceleration is 0.75 m/sec/sec.
On the car only:
Let D = driving force exerted by the car
F = m*a
(D +850) = 1050*(0.75)
D +850 = 787.5
D = 787.5 -850
D = -62.5 N
It is negative. That means the direction of D is opposite that of the resistance to motion of the car. Meaning, D is trying to stop the car, or trying to slow down the car.
So, that is why the acceleration should be negative.

13. Thank you very much for the reply. It makes more sense to me now.

14. Nice work ticbol.

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### car pulling trailer physics problem

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