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Question

Two forces of magnitude F are acting on a uniform disc kept on a rough horizontal surface as shown in the figure. If the frictional force by the horizontal surface on the disc is nF, find the value of n.

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Solution

The force F acting at the centre of the disc will produce zero torque about centre. However, force F acting at the topmost point of the disc will tend to make it rotate in clockwise direction.


Hence, friction will act towards left at the bottom to oppose the tendency of slipping. Let f be the friction force acting on the disc.
Applying Newton's 2nd law on the disc,
2Ff=ma ...(i)
Applying equation of torque about the centre of the disc,
τcom=Icom×α
where τcom= Net torque about centre of mass of the disc.
τN+τmg+(τF)1+(τF)2+τf=Iα
0+0+FR+0+fR=Iα
[τN=τmg=0 because N and mg pass through COM.
Both F and f are tending to rotate disc in clockwise direction, hence the torque will add.]
(F+f)R=Iα ...(ii)

a=Rα, condition for pure rolling & I=mR22
Substituting in Eq.(ii),
(F+f)R=mR22×aR
F+f=ma2...(iii)
Adding Eq.(i) & (iii), we get f=0
Hence according to question,
f=nF or 0=nF
n=0

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