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# M1 Watch

1. A particle P of weight W newtons is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. A horizontal force of magnitude 5N is applied to P. The particle P is in equilibrium with the string taut and with OP making an angle of 25° to the downward vertical.

Edit: Find the tension in the string
2. (Original post by Steelmeat)
A particle P of weight W newtons is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. A horizontal force of magnitude 5N is applied to P. The particle P is in equilibrium with the string taut and with OP making an angle of 25° to the downward vertical.

Edit: Find the tension in the string
Well, what have you tried?
3. Have tried drawing a diagram?
4. (Original post by Zacken)
Well, what have you tried?
so far from P i've drawn a little line with an arrow which has Mg on the end of it

(Original post by ravioliyears)
Have tried drawing a diagram?
Actually the diagram's on the piece of paper i have but i just couldn't be bothered to take a picture and post it on here...
5. (Original post by Steelmeat)
so far from P i've drawn a little line with an arrow which has Mg on the end of it

Actually the diagram's on the piece of paper i have but i just couldn't be bothered to take a picture and post it on here...
I hope the Mg arrow is vertically downwards. Anyhow, resolve horizontally.
6. (Original post by Zacken)
I hope the Mg arrow is vertically downwards. Anyhow, resolve vertically and horizontally.
yup i forgot to mention.

oh ok

Edit: I just seem to be getting this and this for horizontal and vertical

(<----) 5N
Vertically: Mg
7. (Original post by Steelmeat)
yup i forgot to mention.

oh ok

Edit: I just seem to be getting this and this for horizontal and vertical

(<----) 5N
Vertically: Mg
There are components of tension acting both horizonttally and vertically.
8. (Original post by Zacken)
There are components of tension acting both horizonttally and vertically.
I can't seem to see it in the diagram :/
9. (Original post by Steelmeat)
I can't seem to see it in the diagram :/

Tension is an arrow pointing along the string from P to O. It acts at an angle to the downward vertical. Resolve.
10. (Original post by Steelmeat)
I can't seem to see it in the diagram :/
Unsure what downward vertical meant but my attempt -
11. (Original post by Zacken)

Tension is an arrow pointing along the string from P to O. It acts at an angle to the downward vertical. Resolve.
(Original post by Middriver)
Unsure what downward vertical meant but my attempt -
ahhhh i see i just didn't label the tension in the string as a letter oops thanks a lot
12. (Original post by Steelmeat)
ahhhh i see i just didn't label the tension in the string as a letter oops thanks a lot
No worries, is that what downward vertical meant btw?
13. (Original post by Middriver)
No worries, is that what downward vertical meant btw?
yes vertically down is another way of putting it
14. Resolve all the forces in two planes, you will always get a simultaneous or just a simple eq e.g tcos40=5gcos50
15. (Original post by Middriver)
No worries, is that what downward vertical meant btw?
From O, draw a line passing through O, vertically downwards. The angle the string makes with this line is the angle with the downward vertical.
16. (Original post by Zacken)
From O, draw a line passing through O, vertically downwards. The angle the string makes with this line is the angle with the downward vertical.
Thanks
17. (Original post by Steelmeat)
ahhhh i see i just didn't label the tension in the string as a letter oops thanks a lot
(Original post by Middriver)
Thanks
No problem.

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