Ilhajxichjabv
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Question:
Two particles P and Q of equal mass are connected by a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a fixed wedge. One face of the wedge is smooth and inclined to the horizontal at an angle of 30° and the other face of the wedge is rough and inclined to the horizontal at an angle of 60°. Particle P lies on the rough face and particle Q lies on the smooth face with the string connecting them taut. The coefficient of friction between P and the rough face is 0.5.
a Find the acceleration of the system.
b Find the tension in the string.
Solution:
a
(2) ∴
The acceleration of the system is 0.569 m s − 2 ( 3 s.f. ) or 0.57 m s − 2 ( 2 s.f. )
b From (1),
For Q : R (

) , T − m g cos 60 ° = m a
(1)
For P : R ( ր
) , S = m g cos 60 °
R (
ց
) , m g cos 30 ° − S − T 1 2
= m a
m g cos 30 ° − m g cos 60 ° − T 1 2
= m a
( 1 ) + ( 2 ) : m g − \ 3 2
3 m g
4
= 2 ma
( 2\ 3 − 3 )
g
8
= a
cos 30 ° = cos 60 ° = \ 3 2
1
2
T = m g + ( 2\ 3 − 3 ) 1 2 m g 8
= ( 1 + 2\ 3 ) m g 8

I really dont understand this. Can someone explain this please? why does s=mgcos60 and not mgsin60???????????????please been stuck on this for ages.
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brianeverit
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(Original post by Ilhajxichjabv)
Question:
Two particles P and Q of equal mass are connected by a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a fixed wedge. One face of the wedge is smooth and inclined to the horizontal at an angle of 30° and the other face of the wedge is rough and inclined to the horizontal at an angle of 60°. Particle P lies on the rough face and particle Q lies on the smooth face with the string connecting them taut. The coefficient of friction between P and the rough face is 0.5.
a Find the acceleration of the system.
b Find the tension in the string.
Solution:
a
(2) ∴
The acceleration of the system is 0.569 m s − 2 ( 3 s.f. ) or 0.57 m s − 2 ( 2 s.f. )
b From (1),
For Q : R (

) , T − m g cos 60 ° = m a
(1)
For P : R ( ր
) , S = m g cos 60 °
R (
ց
) , m g cos 30 ° − S − T 1 2
= m a
m g cos 30 ° − m g cos 60 ° − T 1 2
= m a
( 1 ) + ( 2 ) : m g − \ 3 2
3 m g
4
= 2 ma
( 2\ 3 − 3 )
g
8
= a
cos 30 ° = cos 60 ° = \ 3 2
1
2
T = m g + ( 2\ 3 − 3 ) 1 2 m g 8
= ( 1 + 2\ 3 ) m g 8

I really dont understand this. Can someone explain this please? why does s=mgcos60 and not mgsin60???????????????please been stuck on this for ages.
Can you edit this so it is legible?
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Ilhajxichjabv
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(Original post by brianeverit)
Can you edit this so it is legible?
no.
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brianeverit
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(Original post by Ilhajxichjabv)
no.
If I can't read it I can't help. Sorry.
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Ilhajxichjabv
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if you know your maths you should be able to work it out. the question is fairly legible. the answer is not.
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Ilhajxichjabv
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davros
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(Original post by Ilhajxichjabv)
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I agree with an answer of 0.5691 for a (taking g=9.8).

S = mgcos60 because this is the component of weight perpendicular to the plane. Along the plane you have a component mgsin60 for P.
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