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#1
My physics exam is tomorow and i did the specimen paper today and im completely stuck on this question. i thought i knew how to do it but what i did was completely different to the ans on the MS

please can someone explain this to me!
thank you
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3 years ago
#2
(Original post by Exotic-L)
My physics exam is tomorow and i did the specimen paper today and im completely stuck on this question. i thought i knew how to do it but what i did was completely different to the ans on the MS

please can someone explain this to me!
thank you
Resolve horizontally, remembering that the tension in each part of the string is the same
0
#3
(Original post by samb1234)
Resolve horizontally, remembering that the tension in each part of the string is the same
could u show me ur working please
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3 years ago
#4
(Original post by Exotic-L)
could u show me ur working please
I'm not going to do the question for you, as you need to be able to do this for yourself in the exam. Think about what component of the tension will be acting in the horizontal direction
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3 years ago
#5
Draw your triangle and think about how the 800 force affects the arrow and THEN put the force into the triangle. Never confuse your self by thinking to much. And could you tell me where this question is from, I would like to try some like these.
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3 years ago
#6
Are you doing AQA. I have physics tomorrow as well, nervous af
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#7
(Original post by moiziqfam)
Draw your triangle and think about how the 800 force affects the arrow and THEN put the force into the triangle. Never confuse your self by thinking to much. And could you tell me where this question is from, I would like to try some like these.
thanks for the reply, im doing WJEC and its from the specimen paper for the newspec. when i do the triangle, to find the resultant force, i would need the vertical component, and i dont know where to put the angle. thats what i dont understand. i think ive been over thinking and now ive just confused myself
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#8
(Original post by samb1234)
Resolve horizontally, remembering that the tension in each part of the string is the same
resolving horizontally is Fcos(70) but i dont understand the tension bit.
i think ive over complicated this and now ive confused myself
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3 years ago
#9
(Original post by Exotic-L)
resolving horizontally is Fcos(70) but i dont understand the tension bit.
i think ive over complicated this and now ive confused myself
We know that the tension in each half of the string is T, and we know that the total horizontal component is 800N. So we can see that 2Tcos70=800
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#10
(Original post by moiziqfam)
Are you doing AQA. I have physics tomorrow as well, nervous af
i think i know im gonna fail, nd ive given up all hope for physics!!
ive done soo much that nothing is going in and i cant understand anything anymore im brain dead atm. its worrying me
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#11
(Original post by samb1234)
We know that the tension in each half of the string is T, and we know that the total horizontal component is 800N. So we can see that 2Tcos70=800
so when a force acts on a string, u think of the string as divided by 2?
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3 years ago
#12
(Original post by Exotic-L)
so when a force acts on a string, u think of the string as divided by 2?
In this case yes
0
3 years ago
#13
(Original post by Exotic-L)
thanks for the reply, im doing WJEC and its from the specimen paper for the newspec. when i do the triangle, to find the resultant force, i would need the vertical component, and i dont know where to put the angle. thats what i dont understand. i think ive been over thinking and now ive just confused myself
You do not need to vertical force. You have the horizontal force. To make things simpler. I will tell u this the forward force is going to be 400 NOT 800 as it is being split into two bow strings. (Newtons third law, every object has equal or opposite reaction). Since the forward force is 400, you know the angle now its basic trig to get the Tension.

if you still dont get the answer only then see my working. I advise you to try it first,.

Forward reaction(horizontal) = 400N
T is the hypotenuse so using trig, T cos 70 = 400, T =1169.52
1
#14
(Original post by samb1234)
In this case yes
sorry to be a pain, but can you explain why we multiply the tension with the cos70? instead of just the 800
0
3 years ago
#15
(Original post by Exotic-L)
sorry to be a pain, but can you explain why we multiply the tension with the cos70? instead of just the 800
From the diagram, forces left=forces right. Therefore 800=2Tcos70
1
#16
(Original post by samb1234)
From the diagram, forces left=forces right. Therefore 800=2Tcos70
thank you soo much! i think i understand it now
0
#17
(Original post by moiziqfam)
You do not need to vertical force. You have the horizontal force. To make things simpler. I will tell u this the forward force is going to be 400 NOT 800 as it is being split into two bow strings. (Newtons third law, every object has equal or opposite reaction). Since the forward force is 400, you know the angle now its basic trig to get the Tension.

if you still dont get the answer only then see my working. I advise you to try it first,.

Forward reaction(horizontal) = 400N
T is the hypotenuse so using trig, T cos 70 = 400, T =1169.52
thank you soo much! i get it now! thank you !!
0
#18
samb1234 and moiziqfam thank you soo much for the help! really appreciate it!
thanks for putting up with me and my stupidity
0
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