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I took moments from a different points to the answer and I am incorrect? A level math

I don't understand what is wrong with my working
Screenshot 2023-05-13 10.04.46 AM.pngScreenshot 2023-05-13 10.04.40 AM.pngIMG_4384.jpg
(edited 11 months ago)
Reply 1
Must admit, its not clear what youve done, but if youve taken moments about B, then you have
clockwise: normal reaction at A
anti clockwise: weight and friction
You seem to have missed out the nomral reaction moment and got the wrong direction for friction moment?
Reply 2
Original post by mqb2766
Must admit, its not clear what youve done, but if youve taken moments about B, then you have
clockwise: normal reaction at A
anti clockwise: weight and friction
You seem to have missed out the nomral reaction moment and got the wrong direction for friction moment?


I have take moments about A

I assumed that the force at B = Frictions

The force at B I have written as Rb
so if I take moments abot A it should be:

Rb x 4sin (alpha) = 40g x cos(alpha)
Reply 3
Original post by leoishush
I have take moments about A

I assumed that the force at B = Frictions

The force at B I have written as Rb
so if I take moments abot A it should be:

Rb x 4sin (alpha) = 40g x cos(alpha)

Hmm, that assumption isnt right. You dont know the angle or magnitude. I guess youve assumed its horziontal which visually it clearly isnt.
Reply 4
Original post by mqb2766
Hmm, that assumption isnt right. You dont know the angle or magnitude. I guess youve assumed its horziontal which visually it clearly isnt.


ahhhhhhhhh i understand. Thank you !
Original post by leoishush
ahhhhhhhhh i understand. Thank you !

Wait, why would you take moments about A if you're trying to find the normal reaction at A? That would get rid of the reaction force from your equation, so you wouldn't be able to calculate it anyway.
Reply 6
Original post by leoishush
I have take moments about A

I assumed that the force at B = Frictions

The force at B I have written as Rb
so if I take moments abot A it should be:

Rb x 4sin (alpha) = 40g x cos(alpha)

re-read the question. Friction is acting at A (as is the normal reaction). You don't know about the force at B so you should be taking moments about B to eliminate it!

Original post by toxicgamage56
Wait, why would you take moments about A if you're trying to find the normal reaction at A? That would get rid of the reaction force from your equation, so you wouldn't be able to calculate it anyway.

This. You can't find a force at a point by taking moments about that point - the effective distance is 0 so the force x distance calc disappears!
Reply 7
Thank you @toxicgamage56 and @davros, I think this is why I often get confused for moments, this will help me massively!
Original post by leoishush
Thank you @toxicgamage56 and @davros, I think this is why I often get confused for moments, this will help me massively!

No problem. If you're edexcel you have ages till stats/mech so dw.
Reply 9
I dont know where ive gone wrong here too, the answer is meant to be tan =2. In the answer they took moments about A
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Reply 10
Original post by leoishush
I dont know where ive gone wrong here too, the answer is meant to be tan =2. In the answer they took moments about A
Screenshot 2023-05-17 9.51.27 AM.jpgScreenshot 2023-05-17 9.39.23 AM.pngScreenshot 2023-05-17 9.53.40 AM.png

Looks like your sin and cos are wrong on the right hand side. Obviously the weight should be the "same trig" as Ra and Fr should be the other.
(edited 11 months ago)
Reply 11
Original post by mqb2766
Looks like your sin and cos are wrong on the right hand side. Obviously the weight should be the "same trig" as Ra and Fr should be the other.


ahhhhhh thank you

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