See the figure given below. $A$ mass of $6 \; kg$ is suspended by a rope of length $2 \; m$ from the ceiling. $A$ force of $50 \; N$ in the horizontal direction is applied at the midpoint $P$ of the rope,as shown. What is the angle the rope makes with the vertical in equilibrium (in $^{\circ}$)? (Take $g = 10 \; m s^{-2}$). Neglect the mass of the rope.

  • A
    $30$
  • B
    $40$
  • C
    $75$
  • D
    $60$

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$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P = 1.9318 \, kg \, wt$,$\sin {\theta _1} = 0.9659$,the value of $R$ is (in $kg \, wt$):

Two masses $m$ and $M$ are attached to the strings as shown in the figure. If the system is in equilibrium,then

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Describe the different types of common forces.

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$A$ particle is situated at the origin of a coordinate system. The following forces begin to act on the particle simultaneously (assuming the particle is initially at rest):
$\vec{F}_1 = 5\hat{i} - 5\hat{j} + 5\hat{k}$
$\vec{F}_2 = 2\hat{i} + 8\hat{j} + 6\hat{k}$
$\vec{F}_3 = -6\hat{i} + 4\hat{j} - 7\hat{k}$
$\vec{F}_4 = -\hat{i} - 3\hat{j} - 2\hat{k}$
Then the particle will move:

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