A block of mass 3 ,kg is pressed against a ve | vedclass

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

Question

(A) The forces acting on the block are as follows:$1$. The normal reaction $N$ exerted by the wall is $N = F \cos 30^{\circ} = F \frac{\sqrt{3}}{2}$.$

Vedclass · Accepted Answer

$30 \,N$

Vedclass · Answer

(A) The forces acting on the block are as follows:$1$. The normal reaction $N$ exerted by the wall is $N = F \cos 30^{\circ} = F \frac{\sqrt{3}}{2}$.$2$. The maximum static friction force is $f_{max} = \mu N = \sqrt{3} 	imes (F \frac{\sqrt{3}}{2}) = \frac{3}{2} F$.$3$. The vertical forces acting on the block are the weight $mg$ acting downwards and the vertical component of the applied force $F \sin 30^{\circ}$ acting downwards.$4$. For the block to be in equilibrium and prevented from falling, the upward friction force must balance the total downward force:$f = mg + F \sin 30^{\circ}$$\frac{3}{2} F = (3 	imes 10) + F 	imes \frac{1}{2}$$\frac{3}{2} F - \frac{1}{2} F = 30$$F = 30 \,N$.

Similar Questions

$A$ marble block of mass $2 \, kg$ lying on ice,when given a velocity of $6 \, m/s$,is stopped by friction in $10 \, s$. The coefficient of friction is:

$A$ body is pulled along a rough horizontal surface with a velocity $6\,m/s$. If the body comes to rest after travelling $9\,m$,then the coefficient of sliding friction is- (Take $g = 10\,m/s^2$)

Consider a car moving along a straight horizontal road with a speed of $72 \, km/h$. If the coefficient of kinetic friction between the tyres and the road is $0.5$,the shortest distance in which the car can be stopped is ........ $m$. $[g = 10 \, m/s^2]$

$A$ vehicle of mass $m$ is moving on a rough horizontal road with momentum $P$. If the coefficient of friction between the tyres and the road is $\mu$,then the stopping distance is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module