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A-Level Maths coefficient of friction question

https://i.imgur.com/ZCS9ntX.png
Have done all the parts was just unsure if i got part c correct.
T= 23.3, Sinx= 3/5 Cosx=4/5
I got
T-2gSinx-F=0
23.3-2g(3/5)=(UR) R=2gCosx R=15.68
23.3-11.76=U(15.68) .. 11.54=U(15.68)
U=11.54/15.68, U=0.736
Reply 1
Original post by ffffffffffffff.
https://i.imgur.com/ZCS9ntX.png
Have done all the parts was just unsure if i got part c correct.
T= 23.3, Sinx= 3/5 Cosx=4/5
I got
T-2gSinx-F=0
23.3-2g(3/5)=(UR) R=2gCosx R=15.68
23.3-11.76=U(15.68) .. 11.54=U(15.68)
U=11.54/15.68, U=0.736

You seem to have assumed its in equilbrium when its actually accelerating at 0.48 as per part a).
Original post by mqb2766
You seem to have assumed its in equilbrium when its actually accelerating at 0.48 as per part a).

oh so is it horizontal= M(0.58)
I assumed as i thought u can only have fmax at equilibrium.
so it should be
T-mgsinx-F=ma
23.3-2g(3/5)-U(15.68)=0.58(2)
23.3-11.76=1.16+U(15.68)
11.54-1.16=UR 10.38=UR
U= 10.38/15.68
U= 0.662?
Reply 3
Original post by ffffffffffffff.
oh so is it horizontal= M(0.58)
I assumed as i thought u can only have fmax at equilibrium.
so it should be
T-mgsinx-F=ma
23.3-2g(3/5)-U(15.68)=0.58(2)
23.3-11.76=1.16+U(15.68)
11.54-1.16=UR 10.38=UR
U= 10.38/15.68
U= 0.662?

You usually assume that during motion (dynamic friction), the frictional force is the limiting value as occurs in static friction. Looks about right, though I think the acceleration is 0.48.
Original post by mqb2766
You usually assume that during motion (dynamic friction), the frictional force is the limiting value as occurs in static friction. Looks about right, though I think the acceleration is 0.48.

thanks

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