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Please see the image attached below to see diagrams for each question:

(1) Given that the string is light and inextensible, and the Tension in the string is 16.06N, find the magnitude of the force exerted on the pulley and the angle that this force makes with the vertical.

(2) Given that the string is light and inextensible, and the Tension in the string is 42.1N, find the magnitude of the force exerted on the pulley.

Also could you PLEASE add a brief explanation to your working...I don't quite understand the method shown for these type of questions in the markschemes.

Thx in advance!

(1) Given that the string is light and inextensible, and the Tension in the string is 16.06N, find the magnitude of the force exerted on the pulley and the angle that this force makes with the vertical.

(2) Given that the string is light and inextensible, and the Tension in the string is 42.1N, find the magnitude of the force exerted on the pulley.

Also could you PLEASE add a brief explanation to your working...I don't quite understand the method shown for these type of questions in the markschemes.

Thx in advance!

C4>O7

The force exerted on the pulley is the net vertical component of the forces in the string either side of the pulley.

So for (1):

F = 16.06sin30 + 16.06sin60

and for (2):

42.1sin30 + 42.1

..whatever they are equal to!

So for (1):

F = 16.06sin30 + 16.06sin60

and for (2):

42.1sin30 + 42.1

..whatever they are equal to!

No mate those answers aren't correct unfortunately...the answers are:

(1) Force on pulley = T*(sqrt2) = 22.7N

which acts at an angle of 15 degrees to the vertical.

(2) Force on pulley = 2Tcos30 = 72.9N

Just to add that the questions are from:

(1) M1 Solomon Paper A

(2) M1 Solomon Paper H

reaction is always halfway, because thats the angle at which the pulley is in contact with the corner.

so look at attachment

pk

so look at attachment

pk

for the force use the (botched to fit) cosine rule...

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

resultant force acts midway between angles from vertical because both forces are the same

(150 - 120)/2 = 15

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

resultant force acts midway between angles from vertical because both forces are the same

(150 - 120)/2 = 15

devesh254

No mate those answers aren't correct unfortunately...the answers are:

(1) Force on pulley = T*(sqrt2) = 22.7N

which acts at an angle of 15 degrees to the vertical.

(2) Force on pulley = 2Tcos30 = 72.9N

Just to add that the questions are from:

(1) M1 Solomon Paper A

(2) M1 Solomon Paper H

(1) Force on pulley = T*(sqrt2) = 22.7N

which acts at an angle of 15 degrees to the vertical.

(2) Force on pulley = 2Tcos30 = 72.9N

Just to add that the questions are from:

(1) M1 Solomon Paper A

(2) M1 Solomon Paper H

Sorry about that- must have been tired

What I suggested was rubbish as it implies that in this situation no force acts on the pulley!

So it's actually the net component in the direction of half the angle between the string either side of the pulley.. I get it now, thanks.

So it's actually the net component in the direction of half the angle between the string either side of the pulley.. I get it now, thanks.

El Stevo

for the force use the (botched to fit) cosine rule...

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

resultant force acts midway between angles from vertical because both forces are the same

(150 - 120)/2 = 15

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

resultant force acts midway between angles from vertical because both forces are the same

(150 - 120)/2 = 15

thx mate...that method seems to work fine! never thought of making it into a triangle and using the cos rule...

Phil23

reaction is always halfway, because thats the angle at which the pulley is in contact with the corner.

so look at attachment

pk

so look at attachment

pk

yeah mate that is the method they have used in the markschemes...i.e. doing 2Tcos30 for (2)...i kinda get it now...you resolve the two T's into two components and use the component which is parallel to R. Then you just resolve the forces parallel to R which would give you 2Tcos45 for (1) and 2Tcos30 for (2)...

lol i can always double check with the botched cos rule el stevo gave me...both work...

thx for ur help everyone!

El Stevo

for the force use the (botched to fit) cosine rule...

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)

[resultant force]^2 = 515.85 - 0

[resultant force] = 22.7 N

nice method - you got to teach me all these weird stuff that you know one day

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