AQA AS 7.2 help

I'm stuck on question 3 which says: an archer pulled a bowstring back until the two halves of the string are at 140 degrees to each other. The force needed to hold the string in this position was 95N. The diagram below it shows a bow with the direction of force opposite to the direction of the arrow and the two halves of the string at 70 degrees to the middle. The top string looks like this: / shape with the other half being it's mirror.
a) calculate the tension in each part of the bowstring in this position
b) calculate the resultant force on an arrow at the instant the bowstring is released

I've tried every method I can think of but none of my answers are the same as in the book, if anyone could get the same answers I'd be very grateful if you showed me how
a=139N b) 95 N
thanks in advance :smile:

Reply 1

roar558

11

use trig for a, the force required to hold the string in place works for both halves of the string so halve it, so use cos70=x/47.5
for part b, if the string releases all it's force on the arrow, this would be 95N, ie the force used to hold it back.

Reply 2

Stonebridge

13

The components, in the direction of the arrow, of the tension in the two strings must add up to 95N for the system to be at equilibrium.
So T cos 70 + Tcos 70 =95N
2T cos 70 = 95
That will give you the answer to a)
The answer to b) should be obvious as it is 95N
This is the force on the arrow from the tension in the two halves of the strings added together, which in part a) we have said is 95N

Reply 3

Megalodon

OP

1

Thanks guys :biggrin:

I kinda got b already but thanks for clarifying it all

Reply 4

fghfhg

5

Original post by roar558

use trig for a, the force required to hold the string in place works for both halves of the string so halve it, so use cos70=x/47.5
for part b, if the string releases all it's force on the arrow, this would be 95N, ie the force used to hold it back.

Ik im 8 years late but im struggling with this question too because i have the exact same texbook and i was wondering why its cosn not sine or toa

Thanks in advance

Reply 5

ailish86

3

a) T1h T2h= 95NT1cos70 = T2Cos70 = 95N(note T1h= T2h)So T1 and T2 = 138.7 = 138N

b) if the force backwards is 95N then the force forward should also be 95N.

(edited 5 years ago)

Reply 6

abdwas

1

Original post by Stonebridge

The components, in the direction of the arrow, of the tension in the two strings must add up to 95N for the system to be at equilibrium.
So T cos 70 + Tcos 70 =95N
2T cos 70 = 95
That will give you the answer to a)
The answer to b) should be obvious as it is 95N
This is the force on the arrow from the tension in the two halves of the strings added together, which in part a) we have said is 95N

where does the cosine come from

Original post by abdwas

where does the cosine come from

It comes from resolving the tension into a horizontal component.

This video would show how force is resolved.

If you still need help, you can watch the video in the spoiler.

Spoiler

AQA AS 7.2 help

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