A block of weight W is held against a vertica | vedclass

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

Question

(C) For the block to be held in equilibrium against the wall,the vertical forces must balance,and the horizontal forces must balance.$1$. Vertical equ

Vedclass · Accepted Answer

Greater than $W$

Vedclass · Answer

(C) For the block to be held in equilibrium against the wall,the vertical forces must balance,and the horizontal forces must balance.$1$. Vertical equilibrium: The frictional force $f$ acting upwards must balance the weight $W$ of the block acting downwards.$f = W$$2$. Horizontal equilibrium: The applied horizontal force $F$ must be balanced by the normal reaction $R$ from the wall.$R = F$$3$. Friction condition: The maximum static frictional force is given by $f_{max} = \mu R$,where $\mu$ is the coefficient of static friction. To hold the block,we need $f \leq \mu R$.Substituting the values from steps $1$ and $2$:$W \leq \mu F$$F \geq \frac{W}{\mu}$Since the coefficient of static friction $\mu$ for most surfaces is less than $1$ $(\mu  W$.Therefore,the minimum force required is $F = \frac{W}{\mu}$,which is greater than $W$.

$A$ block of weight $W$ is held against a vertical wall by applying a horizontal force $F$. The minimum value of $F$ needed to hold the block is

Similar Questions

What does the cell wall of algae consist of?

$A$ plane passes through the point $(3, 5, 7)$. If the direction ratios of its normal are equal to the intercepts made by the plane $x + 3y + 2z = 9$ with the coordinate axes,then the equation of that plane is

An aqueous solution of potash alum gives:

$A$ diverging lens with a focal length of magnitude $25 \, cm$ is placed at a distance of $15 \, cm$ from a converging lens with a focal length of magnitude $20 \, cm$. $A$ beam of parallel light falls on the diverging lens. The final image formed is:

An ideal gas follows a process $PT = \text{constant}$. The correct graph between pressure $(P)$ and volume $(V)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module