A, 5, m long pole of 3, kg mass is placed aga | vedclass

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

Question

Here, $	heta=37^{\circ}$.

Pole is in translational and rotational equilibrium.

So, for translational equilibrium, after balancing the vertical

Vedclass · Accepted Answer

$20\, N$

Vedclass · Answer

Here, $	heta=37^{\circ}$.

Pole is in translational and rotational equilibrium.

So, for translational equilibrium, after balancing the vertical and horizontal forces, $m g=N_2$ and $f=N_1$.

Here, $N_1$ and $N_2$ are normal reaction forces and $f$ is the frictional force.

The balancing torques about $A$ is, $m g \frac{O A}{2}=N_1 O B \quad \ldots$ (i)

(Torque of $N_2$ and $f$ is zero as they are passing from point $A$, so perpendicular distances are zero)

$O A=5 \cos 37^{\circ}=5 \frac{4}{5}=4 m$.

$O B=5 \quad \sin 37^{\circ}=5 \quad \frac{3}{5}=3 m$.

So, from equation (i),

$m g \frac{4}{2}=N_1(3)$

$\Rightarrow N_1=\frac{2}{3}(3)(10)=20\; N$

$m=3\; kg$

Therefore, the frictional force is, $f=N_1=20\; N$.

$A\, 5\, m$ long pole of $3\, kg$ mass is placed against a smooth vertical well as shown in the figure. Under equilibrium condition, if the pole makes an angle of $37^o$ with the horizontal, the frictional force between the pole and horizontal surface is

Similar Questions

Consider the situation shown in the figure. Uniform rod of length $L$ can rotate freely about the hinge $A$ in vertical plane. Pulleys and string are light and frictionless. If therod remains horizontal at rest when the system is released then the mass of the rod is :

Find minimum height of obstacle so that the sphere can stay in equilibrium.

$A$ sphere is placed rotating with its centre initially at rest ina corner as shown in figure $(a)$ & $(b)$. Coefficient of friction between all surfaces and the sphere is $\frac{1}{3}$. Find the ratio of the frictional force $\frac{{{f_a}}}{{{f_b}}}$ by ground in situations $(a)$ & $(b)$.

$A$ weightless rod is acted on by upward parallel forces of $2N$ and $4N$ ends $A$ and $B$ respectively. The total length of the rod $AB = 3m$. To keep the rod in equilibrium a force of $6N$ should act in the following manner: