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A 100kg load is uniformly moved over a horizontal plane by a force F applied at an angle of 30^{\circ} to the horizontal. Find this force if the coefficient of friction between the load and the plane is 0.3. (g=10m/s2)

Option: 1

250N


Option: 2

300N


Option: 3

295.26N


Option: 4

200N


Answers (1)

Given-

mass of the block, m=1000kg

Coefficient of friction, \mu=0.2

Velocity of block is constant (uniform motion).

F.B.D of the block-

As the block is moving with constant velocity (no acceleration) the net force acting on the block must be zero.

\\ \mathrm{\Rightarrow N+Fsin30^0\\} \\ \mathrm{N=mg-\frac{F}{2}}

\\ f=Fcos30^0=\frac{F\sqrt3}{2}\ ...(1)

Kinetic friction-

\\ f=\mu N\\ \Rightarrow f=\mu(mg-\frac{F}{2})\ ...(2)

Comparing equation 1 and 2-

\\ \frac{F\sqrt{3}}{2}=\mu(mg-\frac{F}{2})\\ \\ \Rightarrow F= \frac{2\mu mg}{\sqrt{3}+\mu}=\frac{2000\times0.3}{1.73+0.3}\\ \\ \Rightarrow F=295.26

Posted by

Sumit Saini

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