A body of mass m rests on a horizontal surfac | vedclass

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(C) The forces acting on the body are the gravitational force $mg$ downwards,the normal reaction $R$ upwards,the applied force $P$ at an angle of $30^

Vedclass · Accepted Answer

$\mu \left[ mg - \left( \frac{P}{2} \right) \right]$

Vedclass · Answer

(C) The forces acting on the body are the gravitational force $mg$ downwards,the normal reaction $R$ upwards,the applied force $P$ at an angle of $30^{\circ}$ to the horizontal,and the frictional force $F$ opposing the motion.Resolving the force $P$ into horizontal and vertical components,we get $P \cos 30^{\circ}$ horizontally and $P \sin 30^{\circ}$ vertically upwards.For vertical equilibrium,the sum of upward forces must equal the downward force:$R + P \sin 30^{\circ} = mg$$R = mg - P \sin 30^{\circ}$Since $\sin 30^{\circ} = \frac{1}{2}$,we have:$R = mg - \frac{P}{2}$The limiting friction $F$ is given by $F = \mu R$.Substituting the value of $R$,we get:$F = \mu \left( mg - \frac{P}{2} \right)$.

$A$ body of mass $m$ rests on a horizontal surface. The coefficient of friction between the body and the surface is $\mu$. If the mass is pulled by a force $P$ at an angle of $30^{\circ}$ with the horizontal as shown in the figure,the limiting friction between the body and the surface will be:

Similar Questions

The blocks $A$ and $B$ are arranged as shown in the figure. The pulley is frictionless. The mass of $A$ is $10 \, kg$. The coefficient of friction of $A$ with the horizontal surface is $0.20$. The minimum mass of $B$ to start the motion will be...... $kg$.

For a given mass,how does rolling friction compare to static friction and kinetic friction?

$A$ force $\vec{F}=\hat{i}+4 \hat{j} \, N$ acts on the block of mass $1 \, kg$ shown in the figure. The coefficient of friction between the block and the surface is $\mu = 0.3$. The force of friction acting on the block is (Take $g = 10 \, m/s^2$)

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How does static friction oppose motion? On which factors does the value of the coefficient of friction depend?

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