Someone help pleaseeee

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Sonnysingh22
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#1
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#1
Question: a truck pulls a car up a slope at a constant speed. The truck and the car are joined with a steel tow bar.
The diagram is not drawn to scale.
The lope is 10 degrees to the horizontal ground.
The mass of the car is 1100kg
The car travels from A to B. The vertical distance between A and B is 120m.
There are four forces acting on the car travelling up the slope.

The total frictional force acting on the car as it travels up the slope is 300 N.

Calculate the force provided by the tow bar on the car.

I’m not sure what to do can Someone help
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Sonnysingh22
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#2
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#2
(Original post by Sonnysingh22)
Question: a truck pulls a car up a slope at a constant speed. The truck and the car are joined with a steel tow bar.
The diagram is not drawn to scale.
The lope is 10 degrees to the horizontal ground.
The mass of the car is 1100kg
The car travels from A to B. The vertical distance between A and B is 120m.
There are four forces acting on the car travelling up the slope.

The total frictional force acting on the car as it travels up the slope is 300 N.

Calculate the force provided by the tow bar on the car.

I’m not sure what to do can Someone help
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0le
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#3
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#3
Well first of all draw your free body force diagram and show us how you've attempted to resolve the forces in the relevant directions.
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Sonnysingh22
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#4
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#4
(Original post by 0le)
Well first of all draw your free body force diagram and show us how you've attempted to resolve the forces in the relevant directions.
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Sonnysingh22
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#5
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#5
(Original post by 0le)
Well first of all draw your free body force diagram and show us how you've attempted to resolve the forces in the relevant directions.
And I know that we’re given the frictional force of 300 N so I thought it’d be something to do with this but I’m just not sure how to actually calculate the force. Then I thought you’d use f = ma but I’m not sure because that comes up with an answer of 10791000 which I’m guessing is completely wrong
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medstudent100831
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#6
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#6
looks confusing
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0le
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#7
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#7
(Original post by Sonnysingh22)
[snip]
That’s what I thought it would be
The weight force is drawn incorrectly. It always acts vertically downwards (or more accurately, towards the centre of the Earth).
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Sonnysingh22
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#8
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#8
(Original post by 0le)
The weight force is drawn incorrectly. It always acts vertically downwards (or more accurately, towards the centre of the Earth).
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0le
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#9
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#9
(Original post by Sonnysingh22)
Like that?
Yes. But do you understand why the force has been drawn that way? If not, you should read about Newton's Law of Gravitation.

Moving on, the net force acting on the car has to be zero or in other words, the forces are in equilibrium. This is because the truck is moving at constant speed and you assume the steel bar is rigid such that the car moves at the same speed as the truck from a global reference system such as the ground. This is Newton's First Law. From here you just resolve the forces into their component directions using trigonometry.

The only thing I am wondering about is why they've given you the vertical distance from A to B, unless there is a question about mechanical work.
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Sonnysingh22
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#10
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#10
(Original post by 0le)
Yes. But do you understand why the force has been drawn that way? If not, you should read about Newton's Law of Gravitation.

Moving on, the net force acting on the car has to be zero or in other words, the forces are in equilibrium. This is because the truck is moving at constant speed and you assume the steel bar is rigid such that the car moves at the same speed as the truck from a global reference system such as the ground. This is Newton's First Law. From here you just resolve the forces into their component directions using trigonometry.

The only thing I am wondering about is why they've given you the vertical distance from A to B, unless there is a question about mechanical work.
Yeah because weight always acts downwards?

And so do you mean that the car would not move if it wasn’t for the truck, so the force of the truck acts on the car causing it to move due to the connection of the tow bar?
And I don’t understand how you resolve them with trigonometry, as in I don’t know how to actually start?

And yeah part two of the question asks you to calculate the work done by the force provided by the tow bar as the car travels from A to B. So I’m guessing you use the force we’re trying to work out and the vertical distance to find the force- so that’s why they may have given it?
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0le
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#11
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#11
(Original post by Sonnysingh22)
Yeah because weight always acts downwards?

And so do you mean that the car would not move if it wasn’t for the truck, so the force of the truck acts on the car causing it to move due to the connection of the tow bar?
A force will be transmitted through the tow bar yes. It is a bit complicated to draw out because you have to be careful with Newton's Third Law pairs and take care about what body the forces are acting on.

And I don’t understand how you resolve them with trigonometry, as in I don’t know how to actually start?
With respect, if you don't know how to resolve forces, why are you attempting the question in the first place? You should first of all just learn how to resolve vectors into x and y components or whatever direction you need. In this case, you need to resolve the forces along the direction of the slope.


And yeah part two of the question asks you to calculate the work done by the force provided by the tow bar as the car travels from A to B. So I’m guessing you use the force we’re trying to work out and the vertical distance to find the force- so that’s why they may have given it?
Personally I would have assumed you would calculate the work done along the direction of the force, which in this case is the slope.
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Sonnysingh22
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#12
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#12
(Original post by 0le)
A force will be transmitted through the tow bar yes. It is a bit complicated to draw out because you have to be careful with Newton's Third Law pairs and take care about what body the forces are acting on.


With respect, if you don't know how to resolve forces, why are you attempting the question in the first place? You should first of all just learn how to resolve vectors into x and y components or whatever direction you need. In this case, you need to resolve the forces along the direction of the slope.




Personally I would have assumed you would calculate the work done along the direction of the force, which in this case is the slope.
no I know how to resolve forces but this question i don’t know if it’s the wording, but i just can’t my head around it.

F = mg cos theta

I thought I would use that
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0le
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#13
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#13
(Original post by Sonnysingh22)
no I know how to resolve forces but this question i don’t know if it’s the wording, but i just can’t my head around it.

F = mg cos theta

I thought I would use that
That would give the weight along the direction of the reaction force. You need to calculate the component of weight acting along the direction of friction and then carry out equilibrium of the forces in that direction.
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Sonnysingh22
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#14
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#14
(Original post by 0le)
That would give the weight along the direction of the reaction force. You need to calculate the component of weight acting along the direction of friction and then carry out equilibrium of the forces in that direction.
W sin theta = mg sin theta —> 1100 x 9.81 x sin 10 = 1900N

so 1900 - 300 = 1600N?
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0le
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#15
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#15
(Original post by Sonnysingh22)
W sin theta = mg sin theta —> 1100 x 9.81 x sin 10 = 1900N

so 1900 - 300 = 1600N?
You add the component of weight with the frictional force because they are going in the same direction.
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Muttley79
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#16
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#16
(Original post by Sonnysingh22)
Question: a truck pulls a car up a slope at a constant speed. The truck and the car are joined with a steel tow bar.
The diagram is not drawn to scale.
The lope is 10 degrees to the horizontal ground.
The mass of the car is 1100kg
The car travels from A to B. The vertical distance between A and B is 120m.
There are four forces acting on the car travelling up the slope.

The total frictional force acting on the car as it travels up the slope is 300 N.

Calculate the force provided by the tow bar on the car.

I’m not sure what to do can Someone help
I helped you with this yesterday ... why are you starting another thread.
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Sonnysingh22
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#17
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#17
(Original post by 0le)
you add the component of weight with the frictional force because they are going in the same direction.
1900 +300 = 2200 n?
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Sonnysingh22
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#18
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#18
(Original post by Muttley79)
I helped you with this yesterday ... why are you starting another thread.
I didn’t post anything yesterday? I think you’re mistaking me for someone else
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Muttley79
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#19
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#19
(Original post by Sonnysingh22)
I didn’t post anything yesterday? I think you’re mistaking me for someone else
https://www.thestudentroom.co.uk/sho....php?t=6983666

Hmmm
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0le
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#20
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#20
(Original post by Sonnysingh22)
1900 +300 = 2200 n?
Well Newtons is always a capital letter. So I would expect this is the force in one direction. Therefore you need a force (in the other direction (but still along the same line of action) of 2200 N to achieve an equilibrium along this direction.
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