# Physics Motion QuestionWatch

Thread starter 4 weeks ago
#1

I need help with 11 A and B, don't know how to differentiate the force on the boat and by the horse
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4 weeks ago
#2
(Original post by Tareq Ahmed)

I need help with 11 A and B, don't know how to differentiate the force on the boat and by the horse
Have you tried drawing a diagram?

I don't know if you're familiar with narrowboats and canals, but the towpath where the horse walks is parallel to the canal and a short distance to the side.
The horse pulls the boat along and the boat has to make a slight steering correction away from the side the horse is pulling from.
the angle of the rope to the direction of travel would usually be less than the 40 degrees given by the question, i.e. the horse is a long distance in front of the boat. (not that this will stop us from doing the question)
The rope is virtually horizontal so the force produced at both ends is equal
see https://canalrivertrust.org.uk/refre...1.jpg?v=dfe6fe

https://canalrivertrust.org.uk/refre...3.jpg?v=a5ca95
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4 weeks ago
#3
can you send the final answer???/is it given on that book???
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Thread starter 4 weeks ago
#4
(Original post by cosmos-physics)
can you send the final answer???/is it given on that book???
I don't have the answers yet, I will post as soon as I get them
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4 weeks ago
#5
(Original post by Tareq Ahmed)
I don't have the answers yet, I will post as soon as I get them
hi! bro, is it not given in your book???
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Thread starter 4 weeks ago
#6
(Original post by cosmos-physics)
hi! bro, is it not given in your book???
no, my teacher only have the worksheet itself
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Thread starter 4 weeks ago
#7
(Original post by Tareq Ahmed)
no, my teacher only have the worksheet itself
gave*
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4 weeks ago
#8
(Original post by Tareq Ahmed)
no, my teacher only have the worksheet itself
ok...thanks
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4 weeks ago
#9
(Original post by Tareq Ahmed)

I need help with 11 A and B, don't know how to differentiate the force on the boat and by the horse

This is about applying Newton’s 3rd law

https://www.physicsclassroom.com/cla...on-s-Third-Law

The force exerted by A on B, say FAB is equal and opposite to the force exerted by B on A, say FBA.

FAB = -FBA
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