As shown in the figure,a block of mass sqrt{3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the applied force be $F$. The horizontal component of the force is $F \cos 60^{\circ}$ and the vertical component is $F \sin 60^{\circ}$.The n

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let the applied force be $F$. The horizontal component of the force is $F \cos 60^{\circ}$ and the vertical component is $F \sin 60^{\circ}$.The normal reaction $N$ on the block from the surface is given by balancing vertical forces:$N = mg + F \sin 60^{\circ}$Given $m = \sqrt{3} 	ext{ kg}$ and $g = 10 	ext{ m/s}^2$,so $mg = 10\sqrt{3} 	ext{ N}$.$N = 10\sqrt{3} + F \left( \frac{\sqrt{3}}{2} ight)$The limiting friction force is $f_L = \mu N = \frac{1}{3\sqrt{3}} \left( 10\sqrt{3} + \frac{F\sqrt{3}}{2} ight) = \frac{10}{3} + \frac{F}{6}$.For the block to just move,the horizontal component of the applied force must equal the limiting friction:$F \cos 60^{\circ} = f_L$$F \left( \frac{1}{2} ight) = \frac{10}{3} + \frac{F}{6}$Multiply by $6$ to simplify:$3F = 20 + F$$2F = 20 \Rightarrow F = 10 	ext{ N}$.The question asks for the value of $3x$,where $F = 3x$.$3x = 10$.

Similar Questions

Block $B$ lying on a table weighs $W$. The coefficient of static friction between the block and the table is $\mu$. Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

If the normal force is doubled,the coefficient of friction is

$A$ force of $19.6\, N$ when applied parallel to the surface just moves a body of mass $10\, kg$ kept on a horizontal surface. If a $5\, kg$ mass is kept on the first mass,the force applied parallel to the surface to just move the combined body is........ $N.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module