In the figure, a block of weight 60, N is pla | vedclass

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Question

Frictional force,

${F=\mu R=0.5 	imes m g=0.5 	imes 60=30 \mathrm{N}}$

$\mathrm{Now}$ $\mathrm{F}=\mathrm{T}_{1}=\mathrm{T}_{2} \cos 45^{

Vedclass · Accepted Answer

$30$

Vedclass · Answer

Frictional force,

${F=\mu R=0.5 	imes m g=0.5 	imes 60=30 \mathrm{N}}$

$\mathrm{Now}$ $\mathrm{F}=\mathrm{T}_{1}=\mathrm{T}_{2} \cos 45^{\circ}$

$	ext { or } \quad 30=\mathrm{T}_{2} \cos 45^{\circ}$

${	ext { and }}  {W=T_{2} \sin 45^{\circ}} $

${	herefore \quad}  {W=30 \mathrm{N}}$

In the figure, a block of weight $60\, N$ is placed on a rough surface. The coefficient of friction between the block and the surfaces is $0.5$. ........ $N$ should be the maximum weight $W$ such that the block does not slip on the surface .

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