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JEE Main 2021 (16 Mar, Shift 1) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2021 (16 Mar, Shift 1)

Q26JEE Main 2021 · 16 MAR · SHIFT 1Medium

Torque, Couple & Equilibrium of Rigid Bodies

Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force P⃗\vec{P}of magnitude 100 N is applied at point A of the frame.

Figure

Suppose the force P⃗\vec{P}is resolved parallel to the arms AB and AC of the frame.

The magnitude of the resolved component along the arm AC is xN.

The value of x, to the nearest integer, is ___________.

[Given: sin(35∘^\circ) = 0.573, cos(35∘^\circ) = 0.819

sin(110∘^\circ) = 0.939, cos(110∘^\circ) = −- 0.342]

Numerical answer type. Work it out, then open the answer.

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Answer: 82

  1. The angle at C between rod CA and the wall is 145∘145^\circ, so AC makes 180∘−145∘=35∘180^\circ - 145^\circ = 35^\circ with the vertical.
  2. P⃗\vec{P} is vertical, so the angle between P⃗\vec{P} and arm AC is 35∘35^\circ.
  3. Component along AC =Pcos⁡35∘=100×0.819=81.9= P\cos 35^\circ = 100 \times 0.819 = 81.9 N
  4. So x≈82x \approx 82.
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