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how would i go about doing this question??

https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/June%202014%20QP%20-%20M1%20Edexcel.pdf

question 6

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Reply 1

Zenzation

First I would draw a line on the diagram for the resultant force R which would be in the middle diagonally /. I would also label everything I knew and see where to go from there

Reply 2

Zacken

Original post by thefatone

how would i go about doing this question??

https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/June%202014%20QP%20-%20M1%20Edexcel.pdf

question 6

Resolve and use pythagoras to equate the magnitude to 3X then get a quadratic in X.

Horizontal (in the direction of Q):

(X-20\cos 60^{\circ})

Vertical (upwards):

20\sin 60^{\circ}

(X-20\cos 60^{\circ})^2 + (20 \sin 60^{\circ})^2 = (3X)^2

(edited 8 years ago)

Reply 3

Notnek

Original post by thefatone

how would i go about doing this question??

https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/June%202014%20QP%20-%20M1%20Edexcel.pdf

question 6

You can resolve vertically/horizontally, as shown by Zacken.

Alternatively you could set up a triangle of forces and use the cosine rule. Have you used a triangle of forces before?

Reply 4

thefatone

Original post by notnek

You can resolve vertically/horizontally, as shown by Zacken.

Alternatively you could set up a triangle of forces and use the cosine rule. Have you used a triangle of forces before?

yes but i've forgotten how to do it, show me pls?

Original post by Zacken

Resolve and use pythagoras to equate the magnitude to 3X then get a quadratic in X.

Horizontal (in the direction of Q):

(X-20\cos 60^{\circ})

Vertical (upwards):

20\sin 60^{\circ}

(X-20\cos 60^{\circ})^2 + (20 \sin 60^{\circ})^2 = (3X)^2

i don't understand why X-20cos60

Reply 5

AlmostNotable

Original post by thefatone

yes but i've forgotten how to do it, show me pls?

i don't understand why X-20cos60

There are two forces horizontally. Q:X (right) and P:20cos60 (left), since you are resolving with positive being to the right, you get X-20cos60.

Reply 6

Notnek

Original post by thefatone

yes but i've forgotten how to do it, show me pls?

i don't understand why X-20cos60

You can reposition the vectors to create a triangle of forces:

The red vector R is the resultant force since it's equal to Q + P.

Find angle OQP and then use the cosine rule on triangle OQP to find the magnitude of R.

(edited 8 years ago)

Reply 7

thefatone

Original post by AlmostNotable

There are two forces horizontally. Q:X (right) and P:20cos60 (left), since you are resolving with positive being to the right, you get X-20cos60.

so i have an x going right and a 20cos60 so why x-20cos60? why not add?

Original post by notnek

You can reposition the vectors to create a triangle of forces:

The red vector R is the resultant force since it's equal to Q + P.

Find angle OQP and then use the cosine rule on triangle OQP to find the magnitude of R.

oh ok...

Reply 8

Zacken

Original post by thefatone

so i have an x going right and a 20cos60 so why x-20cos60? why not add?

One is going left and one is going right...

Reply 9

thefatone

Original post by Zacken

One is going left and one is going right...

oh i see.... derp thanks a ton as always for the help :biggrin:

Reply 10

Zacken

Original post by thefatone

oh i see.... derp thanks a ton as always for the help :biggrin:

No problem.

Reply 11

Ayman!

Lami's theorem might be useful to know as Notnek said. :smile:

Reply 12

alesha98

when it says resultant force of P and Q, is it telling me that R is (P+Q), or the 3 forces can be drawn as a closed triangle? and closed triangle means equilibrium? @notnek

Reply 13

Notnek

Original post by alesha98

when it says resultant force of P and Q, is it telling me that R is (P+Q), or the 3 forces can be drawn as a closed triangle? and closed triangle means equilibrium? @notnek

Yes R = P + Q.

And the forces can be drawn as a closed triangle as I showed in my post containing the diagram.

But the forces P and Q which act on the particle are not in equlibrium.

Reply 14

alesha98

Original post by notnek

Yes R = P + Q.

And the forces can be drawn as a closed triangle as I showed in my post containing the diagram.

But the forces P and Q which act on the particle are not in equlibrium.

What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

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Reply 15

Notnek

Original post by alesha98

What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

Posted from TSR Mobile

If you have 3 vectors a, b and c such that a = b + c then by defintion of vector addition, you can draw the vectors in a closed triangle.

Think back to GCSE vectors.

I'll post a diagram for 6b - give me a minute.

Reply 16

alesha98

Original post by notnek

Ahh now I get it. This is very useful to me ... I completely forgot that ... thankyou so much

Posted from TSR Mobile

Reply 17

alesha98

Original post by alesha98

Ahh now I get it. This is very useful to me ... I completely forgot that ... thankyou so much

Posted from TSR Mobile

Takes your time to draw 6b. I will wait ,thanks

Posted from TSR Mobile

Reply 18

Notnek

Original post by alesha98

What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

Posted from TSR Mobile