A man balances himself in a horizontal positi | vedclass

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

Question

(B) Let the normal force exerted by the man on each wall be $N$. Since the man is in equilibrium,the total horizontal force must be zero,so the walls

Vedclass · Accepted Answer

He exerts only horizontal forces on the walls

Vedclass · Answer

(B) Let the normal force exerted by the man on each wall be $N$. Since the man is in equilibrium,the total horizontal force must be zero,so the walls exert an equal normal force $N$ on the man.Let the friction force exerted by each wall on the man be $f$. For vertical equilibrium,$2f = mg$,so $f = mg/2$. Thus,the friction forces are equal.The total force exerted by each wall on the man is the vector sum of the normal force $N$ and the friction force $f$,which is $\vec{R} = \vec{N} + \vec{f}$. Since $\vec{f}$ is vertical and $\vec{N}$ is horizontal,the resultant force $\vec{R}$ is not horizontal.Option $(B)$ states that he exerts only horizontal forces on the walls. However,to maintain equilibrium,the man must also exert a downward frictional force on the walls to balance the upward frictional force exerted by the walls on him. Therefore,he exerts both horizontal and vertical forces on the walls. Thus,$(B)$ is incorrect.

$A$ man balances himself in a horizontal position by pushing his hands and feet against two parallel walls. His centre of mass lies midway between the walls. The coefficients of friction at the walls are equal. Which of the following is not correct?

Similar Questions

$A$ block $B$,lying on a table,has weight $w$. 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

$A$ man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground,and between the block and the ground.

Given $m_1 = 4m_2$. The acceleration of $m_1$ is $a$. Find the velocity of $m_2$ at $t = 0.4 \, s$ in $cm/s$.

$A$ body of mass $2 \ kg$ is initially at rest. It moves on a table under the influence of a horizontal force of $7 \ N$. If the coefficient of kinetic friction is $0.1$,then the work done by the applied force and the work done by the frictional force in $10 \ s$ are respectively:

An object of mass $10 \,kg$ is projected from the ground with a speed of $40 \,m/s$ at an angle of $60^{\circ}$ with the horizontal. The rate of change of momentum of the object one second after projection in $SI$ units is ........ [Take $g = 9.8 \,m/s^2$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module