# OCR M1 Retake Help

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#1
Hi guys, Im not sure how to do q.8 in Misc. Ex4, p.60 in the OCR M1 book. I would really appreciate somebody telling me how to do the question. I do know how to resolve and have gotten all the other ones in the Misc Ex right , but the equations I end up with are:

1) 1800cos20 + 1650 cosx - F ( resistive force) = 0.6 a ( R (horizontally))
2) 39200N ( weight 4000X 9.8) = R + 1650sinx + 1800sin20 ( R ( vertically))

In order to find out x, I know that I need to use equation 2, but there are too many variables in eq.2 to find out x, because I need to include R ( normal contact force).

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10 years ago
#2
(Original post by Arian315)
Hi guys, Im not sure how to do q.8 in Misc. Ex4, p.60 in the OCR M1 book. I would really appreciate somebody telling me how to do the question. I do know how to resolve and have gotten all the other ones in the Misc Ex right , but the equations I end up with are:

1) 1800cos20 + 1650 cosx - F ( resistive force) = 0.6 a ( R (horizontally))
2) 39200N ( weight 4000X 9.8) = R + 1650sinx + 1800sin20 ( R ( vertically))

In order to find out x, I know that I need to use equation 2, but there are too many variables in eq.2 to find out x, because I need to include R ( normal contact force).

Post the question. Not many people may have that textbook or, like me, may do but cannot be bothered to go and look for it.
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#3
Ok - here is the question:

A barge of mass 4 tonnes is pulled in a a straight line by two tugs with an acceleration of 0.6m/s2. The tension in one two rope is 1800N, and the tension in the other is 1650N. Given that the angles the ropes make with the direction of motion are 20 degrees and x degrees respectively, find the angle x and the resistance to motion of the barge.

The equations I end up with are:

1) 1800cos20 + 1650 cosx - F ( resistive force) = 0.6 a ( R (horizontally))
2) 39200N ( weight 4000X 9.8) = R + 1650sinx + 1800sin20 ( R ( vertically))

In order to find out x, I know that I need to use equation 2, but there are too many variables in eq.2 to find out x, because I need to include R ( normal contact force).
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10 years ago
#4
(Original post by Arian315)
Ok - here is the question:

A barge of mass 4 tonnes is pulled in a a straight line by two tugs with an acceleration of 0.6m/s2. The tension in one two rope is 1800N, and the tension in the other is 1650N. Given that the angles the ropes make with the direction of motion are 20 degrees and x degrees respectively, find the angle x and the resistance to motion of the barge.

The equations I end up with are:

1) 1800cos20 + 1650 cosx - F ( resistive force) = 0.6 a ( R (horizontally))
2) 39200N ( weight 4000X 9.8) = R + 1650sinx + 1800sin20 ( R ( vertically))

In order to find out x, I know that I need to use equation 2, but there are too many variables in eq.2 to find out x, because I need to include R ( normal contact force).
You don't need the normal contact force. Just resolve the forces parallel and perpendicular to the direction of motion. (Both of these directions are horizontal).

The first equation you have is correct, you just need an equation for the forces perpendicular to the direction of motion. The second equation should only have one unknown (the angle x)
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#5
Wait but why don't you need the normal contact force? I was always taught that if an object is on a surface, there is always a normal contact force....or is this to do with water or something??
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10 years ago
#6
(Original post by Arian315)
Wait but why don't you need the normal contact force? I was always taught that if an object is on a surface, there is always a normal contact force....or is this to do with water or something??
There is a normal contact force but you don't need it since it is equal to the weight of the barge (since there is no overall vertical force)
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#7
But in the previous question I did, if there was a string on an object on a slope, it has a vertical effect on the object. Therefore the normal contact force is reduced. E.G. if there is a string with tension 10N at 10 degrees to the horizontal on an object of weight 100 N on a flat surface, then :

10sin10 + R = W ( 100N)

R= 100- 10sin10 - not just 100 right?

Sorry, I hope you don't think Im arguing with you. I just want to make sure I completely understand what you are saying so that I can get it right in the exam. Thanks.
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10 years ago
#8
(Original post by Arian315)
But in the previous question I did, if there was a string on an object on a slope, it has a vertical effect on the object. Therefore the normal contact force is reduced. E.G. if there is a string with tension 10N at 10 degrees to the horizontal on an object of weight 100 N on a flat surface, then :

10sin10 + R = W ( 100N)

R= 100- 10sin10 - not just 100 right?

Sorry, I hope you don't think Im arguing with you. I just want to make sure I completely understand what you are saying so that I can get it right in the exam. Thanks.
The reason you have to take into account the reaction force in that example is because the tension in the string has a vertical component. In the example of the barge and the tugs, both of the tensions and the resistive force are all horizantal and have no vertical component which means that they don't have any impact on the reaction force.

You might find the question easier if you have a diagram to look at.
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#9
But the tugs are at 20 degrees ( 1800N) and x degrees ( 1650N) to the horizontal. How can they not have a vertical component? ( should be 1800sin20 and 1650sin x right?). And yeah if I could draw what I was thinking on TSR I would...
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10 years ago
#10
(Original post by Arian315)
But the tugs are at 20 degrees ( 1800N) and x degrees ( 1650N) to the horizontal. How can they not have a vertical component? ( should be 1800sin20 and 1650sin x right?). And yeah if I could draw what I was thinking on TSR I would...
The tugs are at 20 degrees and x degrees to the direction of motion but are still horizontal. Think of it as both tugs are in front of the boat but one is on the left and the other is on the right.
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#11
So what would the equations become then? This?:

1) 1800cos20 + 1650 cosx - F ( resistive force) = 0.6 a ( R (horizontally))
2) 39200N ( weight 4000X 9.8) = 1650sinx + 1800sin20 ( R ( vertically)) ( ignoring R)

then you can find out x and then sub it into (1) to find R, which is what the question asks
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10 years ago
#12

This is the best diagram I can do. It shows a horizontal plane with the vertical direction going into/coming out of the page. The dotted arrow represents the direction of motion whilst the others show forces.

EDIT: the diagram shows all of the forces you need. you don't need weight or reaction
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#13
Thanks a lot...yeah it doesn't say in the question whether that 20 degrees is below the horizontal or not, cos I drew them both as above the horizontal ( if you know what I mean). If they had drawn a diagram like that it would have made more sense...
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5 years ago
#14
1650sinx = 1800sin20

(1800sin20)/1650 = 0.373

Sin-1(0.373) = 21.9

x = 21.9

Look at the diagram as if from Birdseye view, both tugs are flat, completely horizontal.

R = resistance to motion of barge

1800cos20 + 1650cos21.9 - R = 2400

3222.376 - R = 2400

R = 822.3N

The question is pretty badly phrased.
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