A block of mass 15 , kg is resting on a rough | vedclass

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

Question

(A) The block is in equilibrium,so the net force acting on it is zero. We consider the forces along the inclined plane ($x$-axis) and perpendicular to

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) The block is in equilibrium,so the net force acting on it is zero. We consider the forces along the inclined plane ($x$-axis) and perpendicular to it ($y$-axis).The forces acting on the block are:$1$. Weight $mg = 15 	imes 10 = 150 \, N$ acting vertically downwards.$2$. Normal reaction $N$ perpendicular to the plane.$3$. Tension $T = 50 \, N$ acting horizontally.$4$. Limiting friction $f = \mu N$ acting up the plane.Resolving forces along the $x$-axis (parallel to the plane):$\Sigma F_x = 0$$f + T \cos 45^{\circ} = mg \sin 45^{\circ}$$\mu N + 50 \cos 45^{\circ} = 150 \sin 45^{\circ} \quad ...(i)$Resolving forces along the $y$-axis (perpendicular to the plane):$\Sigma F_y = 0$$N = mg \cos 45^{\circ} + T \sin 45^{\circ}$$N = 150 \cos 45^{\circ} + 50 \sin 45^{\circ} \quad ...(ii)$Substituting $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ in equation $(ii)$:$N = (150 + 50) 	imes \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, N$Substituting $N$ into equation $(i)$:$\mu (100\sqrt{2}) + 50 	imes \frac{1}{\sqrt{2}} = 150 	imes \frac{1}{\sqrt{2}}$$\mu (100\sqrt{2}) = \frac{100}{\sqrt{2}}$$\mu = \frac{100}{\sqrt{2} 	imes 100\sqrt{2}} = \frac{1}{2}$

$A$ block of mass $15 \, kg$ is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal string which has a tension of $50 \, N$. The minimum coefficient of friction between the surfaces of contact is $(g = 10 \, m/s^2)$.

Similar Questions

When a body is placed on a rough plane (coefficient of friction $= \mu$) inclined at an angle $\theta$ to the horizontal,its acceleration is (acceleration due to gravity $= g$)

$A$ block is projected upwards on an inclined plane of inclination $37^{\circ}$ along the line of greatest slope with a coefficient of friction $\mu = 0.5$ and an initial velocity of $5 \ m/s$. The block first stops at a distance of .......... $m$ from the starting point.

$A$ body of mass $100\, g$ is sliding down an inclined plane of inclination $30^\circ$. What is the frictional force experienced if the coefficient of friction $\mu = 1.7$?

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F)$ required to move the block up the inclined plane will be

$A$ uniform rope of length $l$ lies on a table. If the coefficient of friction is $\mu$,then the maximum length $l_1$ of the part of this rope which can overhang from the edge of the table without sliding down is

Vedclass Test Series

Exam Paper Generator

Online Exam Module