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###### A-Level Maths coefficient of friction question

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2 months ago

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

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

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

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).

Reply 2

2 months ago

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?

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?

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.

Reply 4

2 months ago

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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