Question

Paragraph for Question Nos. 35 to 37 The block A of mass 10 kg is kept on platform P of mass 25 kg. A force of 25 N is applied on P and force of 10N is applied on A as shown in the figure. Friction is absent between platform and ground. Coefficient of friction between platform and block is µ = 0.06.

35. Direction of pseudo force on A as seen from P will be at an angle $\theta$ from negative x-axis where

• A. $\theta$ = 0°
• B. $\theta$ <45°
• C. $\theta$ = 45°
• D. $\theta$ >45°

Answer 1

Answer: (D)

$\theta$ >45°

Answer 2

The acceleration of platform P is $\vec{a}_P = a_{Px} \hat{i} + a_{Py} \hat{j}$. $a_{Px} = -1 + 0.24 \cos 45^\circ \approx -0.8304 \, \text{m/s}^2$. $a_{Py} = 0.24 \sin 45^\circ \approx 0.1696 \, \text{m/s}^2$. The pseudo force on A as seen from P is $\vec{F}_{pseudo} = -m_A \vec{a}_P = -m_A (a_{Px} \hat{i} + a_{Py} \hat{j})$. $\vec{F}_{pseudo} = -10 (-0.8304 \hat{i} + 0.1696 \hat{j}) = (8.304 \, \text{N}) \hat{i} - (1.696 \, \text{N}) \hat{j}$. This force vector is in the 4th quadrant (positive x-component, negative y-component). Let $\phi$ be the angle of $\vec{F}_{pseudo}$ with the positive x-axis. $\tan \phi = \frac{-1.696}{8.304} \approx -0.204$. $\phi \approx -11.5^\circ$. (This is $11.5^\circ$ clockwise from the positive x-axis). The question asks for the angle $\theta$ from the negative x-axis. The negative x-axis is at $180^\circ$ from the positive x-axis. The angle $\theta$ can be found as the angle between the vector $\vec{F}_{pseudo}$ and the vector $(-\hat{i})$. $\cos \theta = \frac{\vec{F}_{pseudo} \cdot (-\hat{i})}{|\vec{F}_{pseudo}| \cdot |-\hat{i}|} = \frac{(8.304)( -1) + (-1.696)(0)}{\sqrt{8.304^2 + (-1.696)^2} \times 1} = \frac{-8.304}{\sqrt{68.95 + 2.87}} = \frac{-8.304}{\sqrt{71.82}} \approx \frac{-8.304}{8.474} \approx -0.9799$. $\theta = \arccos(-0.9799) \approx 168.5^\circ$. Since $168.5^\circ > 45^\circ$, the correct option is (D).