Question

Direction of psuedo force on $A$ as seen from $P$ will be at angle $\theta$ from negative $x$-axis where:

• A. $\theta =0^o$
• B. $\theta <{45}^o$
• C. $\theta ={45}^o$
• D. $\theta >{45}^o$

Answer 1

The correct option is B $\theta <{45}^o$

$N_1=M_{A}g$

acceleration along x-axis;

$a_x=\frac{F}{M+M_A}=\frac{25}{25+10}=\frac{25}{35}=\frac{5}{7}$

acceleration along y-axis;

$a_y=\frac{F}{25}=\frac{\mu \left(10g\right)}{25}=\frac{0.06\times 10\times 10}{25}=\frac{6}{25}$

Mow, pseudo force on block A is shown in fig.

Resultant force of pseudo force is $P=m_a{a}_y\left(\widehat{-j}\right)+m_a{a}_x\left(\hat{i}\right)$

Angle, $\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$

$\theta ={\tan }^{-1}\left(\frac{\frac{6}{25}}{\frac{5}{7}}\right)$

$\theta ={\tan }^{-1}\left(\frac{42}{125}\right)<{45}^{\circ }$