# MomentsWatch

#1

How do i do part B?
ignore all working
diagram of system is attached

give me an answer and talk me through how you did it.
(i have a copy of the teachers answer here anyway so trying to get me to work it out is useless since i can't do it)
Spoiler:
Show
Just realised you can see part of my finger o.o
0
2 years ago
#2
(Original post by thefatone)

How do i do part B?
ignore all working
diagram of system is attached

give me an answer and talk me through how you did it.
(i have a copy of the teachers answer here anyway so trying to get me to work it out is useless since i can't do it)
Spoiler:
Show
Just realised you can see part of my finger o.o
The key here is that that you're applying a moment to the end of the rod using the 2kg mass on the pulley. The mass is fixed, but you can move where the pulley is, so where do you need to move it so that the moment is greatest?
0
2 years ago
#3
(Original post by thefatone)

How do i do part B?
ignore all working
diagram of system is attached

give me an answer and talk me through how you did it.
(i have a copy of the teachers answer here anyway so trying to get me to work it out is useless since i can't do it)
Spoiler:
Show
Just realised you can see part of my finger o.o
Take moments around B. We know that the force perpendicular to the rod provided by the string will be Tcos theta, so the max will be when theta is 0 so it will equal T. We can then use your calculated value for T, and the data in the question to work it out (since for equilibrium the moment about any point is 0)
0
#4
(Original post by lerjj)
The key here is that that you're applying a moment to the end of the rod using the 2kg mass on the pulley. The mass is fixed, but you can move where the pulley is, so where do you need to move it so that the moment is greatest?
*shutdown*...........

................................ ... to make the moment greatest you should move the pulley horizontally so the angle becomes 90° when this happens there's no horizontal component so all the force is just vertical?

(Original post by samb1234)
Take moments around B. We know that the force perpendicular to the rod provided by the string will be Tcos theta, so the max will be when theta is 0 so it will equal T. We can then use your calculated value for T, and the data in the question to work it out (since for equilibrium the moment about any point is 0)
oh ffs thanks for making me see this.....
0
#5
(Original post by samb1234)
Take moments around B. We know that the force perpendicular to the rod provided by the string will be Tcos theta, so the max will be when theta is 0 so it will equal T. We can then use your calculated value for T, and the data in the question to work it out (since for equilibrium the moment about any point is 0)
wait a second i got 2x1.6=mgx1.2 + 3.5x1.8

why is the mass m not mg? why is simply just m?

surely gravity acts upon it?
0
2 years ago
#6
(Original post by thefatone)
wait a second i got 2x1.6=mgx1.2 + 3.5x1.8

why is the mass m not mg? why is simply just m?

surely gravity acts upon it?
sounds like they/you have cancelled g, since the forces are all weights so g can be cancelled from everything
0
#7
(Original post by samb1234)
sounds like they/you have cancelled g, since the forces are all weights so g can be cancelled from everything
oh.... ,-, i didn't see that
0
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