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M1 doubts

Ruki

Ruki

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Ruki

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robotic94 said:
and ii) T – 0.227 × 5 = 0.5a

why 0.227 × 5?
Click to expand...

0.227 x 5 comes from the equation: F = µR

So to find the friction we use the coefficient of friction we found from the previous answer 0.227 and the R which is 0.5g which is 5N from 0.5*10. We dont have a 3.2sin30 in this case since they take if off for this question.

Hope this helps (y)
 
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Nibz

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Untitled.png Since Q hangs vertically below the pulley, the angle it makes with the straight horizontal string is 90.

2 T cos45 = 4 √2
T√2 = 4√2
T = 4N
 
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1357913579 said:
Thanks alot:),
one more question please same paper number 7 second part
Click to expand...
Np!

When t= 2.5 s
d = 2.6 (2.5) = 6.5 m

Since the vertical height b/w P and Q = 1.6m
Draw a triangle.
sin theta = (vertical height) / hypotenuse (distance)
Sin theta = 1.6 / 6.5
Sin theta = 0.2462

From part (i) you get the equation for acceleration to be a = 10 sin theta
so a = 10 (0.2462) = 2.46 m/s^2
 
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1357913579

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Nibz said:
Np!

When t= 2.5 s
d = 2.6 (2.5) = 6.5 m

Since the vertical height b/w P and Q = 1.6m
Draw a triangle.
sin theta = (vertical height) / hypotenuse (distance)
Sin theta = 1.6 / 6.5
Sin theta = 0.2462
thanks once again:)
but i didnt got 19 sin theta from first part. do we get it like g sintheta ?

From part (i) you get the equation for acceleration to be a = 10 sin theta
so a = 10 (0.2462) = 2.46 m/s^2
Click to expand...
 
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donorsolutions

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blabla said:
Q:3 http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_41.pdf
somebody help pleeeease! :(
Click to expand...
Tension in AP1 is 5.5 N, and in CP2 is WN. Because the particles are in equilibrium , TENSION= Weight.
Now because they are attached with a smooth pulley, P1X will also have a tension of 5.5 N. Similarly, P2X will have a tension of W N. I hope things are clear till this extent.
Now consider the point X, it has 2 tensions of 5.5 N and W N acting away from it and there is another particle B (Weight 7.3N) attached to the point X.
NOW NOTICE Angle P1XP2 is right angle!
so because there is no resultant of forces on the point X (because in equilibrium). The resultant of the two tensions 5.5 and W should be 7.3, thats how X remains in equilibrium and balances weight of B.
So (5.5)^2 + (W)^2 = (7.3)^2
W comes out to be 4.8 N.
Now draw a vertical straight line at X. Notice that AP1 is parallel to your drawn line.So the angle AP1X (let suppose alpha) would be equal to the angle subtended at the left portion of X. The right portion of X then has an angle (90-alpha). Now resolve forces horizontally.

5.5sinaplha=4.8sin(90-aplha)
Remeber the identity sin(90-aplha) = cos alpha
cos(90-alpha) = sin alpha

So Equation Becomes :
5.5 sinalpha=4.8 cosalpha
tan alpha = 4.8/5.5
take out the value of alpha..phew hope that helps. Waiting for your reply.
 
blabla

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donorsolutions said:
Tension in AP1 is 5.5 N, and in CP2 is WN. Because the particles are in equilibrium , TENSION= Weight.
Now because they are attached with a smooth pulley, P1X will also have a tension of 5.5 N. Similarly, P2X will have a tension of W N. I hope things are clear till this extent.
Now consider the point X, it has 2 tensions of 5.5 N and W N acting away from it and there is another particle B (Weight 7.3N) attached to the point X.
NOW NOTICE Angle P1XP2 is right angle!
so because there is no resultant of forces on the point X (because in equilibrium). The resultant of the two tensions 5.5 and W should be 7.3, thats how X remains in equilibrium and balances weight of B.
So (5.5)^2 + (W)^2 = (7.3)^2
W comes out to be 4.8 N.
Now draw a vertical straight line at X. Notice that AP1 is parallel to your drawn line.So the angle AP1X (let suppose alpha) would be equal to the angle subtended at the left portion of X. The right portion of X then has an angle (90-alpha). Now resolve forces horizontally.

5.5sinaplha=4.8sin(90-aplha)
Remeber the identity sin(90-aplha) = cos alpha
cos(90-alpha) = sin alpha

So Equation Becomes :
5.5 sinalpha=4.8 cosalpha
tan alpha = 4.8/5.5
take out the value of alpha..phew hope that helps. Waiting for your reply.
Click to expand...

thank you so much! that was a very good explanation :)
 
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donorsolutions

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aditya avhad said:
Oh..but how is it 0.8m..im getting 0.4 :(
Click to expand...
Remember the question asks :
find the total distance
travelled by Q from the instant it first reaches X until it returns to X.
hahah man you have done 99% of that question..but failed to obtain the right answer. The distance you calculated 0.4 metres is the distance the particle travels from X to its highest point. After that it will fall again, to the point X.(question demands of the time taken from leaving X and then coming back to X again.) So we will have to multiply it by 2.
So 2(0.4)= 0.8 metres.
Hope that Helps :D
 
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Nibz

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iKhaled said:
can i pls know y is it " T Cos 45 " :S:S:S:S:S:S
Click to expand...
Because Q hangs 'VERTICALLY' below the pulley which means that the angle that it makes with pulley is 90. Horizontal component acts right in the middle, so the angle would be 45 and hence T cos 45.
 
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