A mass of 5, kg is suspended by a rope of len | vedclass

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

Question

(D) Let the mass be $m = 5\, kg$. The weight of the mass is $W = mg = 5 	imes 10 = 50\, N$ (taking $g = 10\, m/s^2$).When a horizontal force $F = 50\

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Let the mass be $m = 5\, kg$. The weight of the mass is $W = mg = 5 	imes 10 = 50\, N$ (taking $g = 10\, m/s^2$).When a horizontal force $F = 50\, N$ is applied at the mid-point,the rope deflects by an angle $	heta$ with the vertical.At equilibrium,the forces acting on the point where the force is applied are:$1$. The horizontal force $F = 50\, N$ acting to the right.$2$. The tension component $T \sin 	heta$ acting to the left.$3$. The tension component $T \cos 	heta$ acting upwards.$4$. The weight $W = 50\, N$ acting downwards.For equilibrium in the horizontal direction: $T \sin 	heta = F = 50\, N$.For equilibrium in the vertical direction: $T \cos 	heta = W = 50\, N$.Dividing the two equations: $\frac{T \sin 	heta}{T \cos 	heta} = \frac{50}{50} \Rightarrow 	an 	heta = 1$.Therefore,$	heta = 	an^{-1}(1) = 45^o$.

$A$ mass of $5\, kg$ is suspended by a rope of length $2\, m$ from a ceiling. $A$ force of $50\, N$ in the horizontal direction is applied at the mid-point of the rope. The angle made by the rope with the vertical,in equilibrium is ........ $^o$

Similar Questions

Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also,name the triangle formed by the forces as sides.

If a body is in equilibrium under a set of non-collinear forces,then the minimum number of forces has to be

$A$ mass '$M$' is suspended by a rope from a rigid support at point '$P$' as shown in the figure. Another rope is tied at end '$Q$' and pulled horizontally with a force '$F$'. If the rope makes an angle '$\theta$' with the vertical,then the tension in the string '$PQ$' is

$A$ spherical ball of mass $m=5\,kg$ rests between two planes which make angles of $30^{\circ}$ and $45^{\circ}$ respectively with the horizontal. The system is in equilibrium. Find the normal forces exerted on the ball by each of the planes. The planes are smooth.

As shown in the figure,a $70\,kg$ garden roller is pushed with a force of $\overrightarrow{F} = 200\,N$ at an angle of $30^{\circ}$ with the horizontal. The normal reaction on the roller is $.......\,N$ (Given $g = 10\,m s^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module