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Question: In the system shown in figure, all the surfaces are smooth. Rod is moved by external agent with acce...

In the system shown in figure, all the surfaces are smooth. Rod is moved by external agent with acceleration 9 m/s² vertically downwards. Force exerted on the rod by the wedge will be:

A

120 N

B

200 N

C

135/2 N

D

225/2 N

Answer

200 N

Explanation

Solution

Solution

  1. Setup and Geometry

The rod is accelerated vertically downward with acceleration a⃗r=(0,−9)\vec{a}_r=(0,-9) m/s². The contact is on the wedge’s smooth inclined face which makes an angle 37∘37^\circ with the horizontal. Choosing the unit normal to the inclined face as

n^=(−sin⁡37∘,cos⁡37∘),\hat{n}=(-\sin37^\circ,\cos37^\circ),

(pointing from the wedge toward the rod), the contact condition (no separation along the normal) requires that the relative acceleration along n^\hat{n} must vanish.

  1. Relative Acceleration Condition

Let the wedge accelerate horizontally with unknown a⃗w=(a,0)\vec{a}_w=(a,0). Then the relative acceleration is:

a⃗rel=a⃗r−a⃗w=(−a, −9).\vec{a}_{\text{rel}}=\vec{a}_{r}-\vec{a}_w=( -a,\,-9).

The no‐separation condition along n^\hat{n} gives:

a⃗rel⋅n^=0⟹(−a, −9)⋅(−sin⁡37∘,cos⁡37∘)=0.\vec{a}_{\text{rel}}\cdot\hat{n}=0\quad\Longrightarrow\quad (-a,\,-9)\cdot(-\sin37^\circ,\cos37^\circ)=0.

This yields:

asin⁡37∘−9cos⁡37∘=0⟹a=9cos⁡37∘sin⁡37∘=9cot⁡37∘.a\sin37^\circ - 9\cos37^\circ = 0\quad\Longrightarrow\quad a = \frac{9\cos37^\circ}{\sin37^\circ}=9\cot37^\circ.

Numerically, using sin⁡37∘≈0.6018\sin37^\circ\approx 0.6018 and cos⁡37∘≈0.7986\cos37^\circ\approx 0.7986,

a≈9×0.79860.6018≈9×1.327≈11.94 m/s2.a\approx 9\times\frac{0.7986}{0.6018}\approx 9\times1.327\approx 11.94\ \text{m/s}^2.
  1. Wedge Dynamics

The only horizontal force acting on the 10 kg wedge is due to the reaction force NN from the rod. By Newton’s third law, the wedge experiences a force:

F⃗w=−Nn^.\vec{F}_w=-N\hat{n}.

Only the horizontal component matters. Since

Nn^=N(−sin⁡37∘,cos⁡37∘),N\hat{n}=N(-\sin37^\circ,\cos37^\circ),

the horizontal component is Nsin⁡37∘N\sin37^\circ (directed to the right). Applying Newton’s second law for the wedge in the horizontal direction:

10 a=Nsin⁡37∘.10\,a = N\sin37^\circ.

So,

N=10 asin⁡37∘=10×11.940.6018≈119.40.6018≈198.3 N.N = \frac{10\,a}{\sin37^\circ} = \frac{10\times 11.94}{0.6018}\approx \frac{119.4}{0.6018}\approx 198.3\ \text{N}.

Rounding, the force NN is approximately 200 N.