Question

Two blocks connected by a massless string slide on an inclined plane having an angle of inclination of ${37}^{\circ }$. The masses of the two blocks are $M_1=4kg$ and $M_2=2kg$ respectively and the coefficients of friction of $M_1$ and $M_2$ with the inclined plane are $0.75$ and $0.25$ respectively. Assuming the string to be taut, find (i) the common acceleration of two masses and (ii) the tension in the string. $(sin{37}^{\circ }=0.6,cos{37}^{\circ }=0.8)$ [Take $g=9.8m{s}^{-2}$]

• A. $a=2.3m{s}^{-2},T=8.45N$
• B. $a=1.3m{s}^{-2},T=5.24N$
• C. $a=1.35m{s}^{-2},T=11.0N$
• D. $a=9.4m{s}^{-2},T=12.5N$

Answer 1

The correct option is B $a=1.3m{s}^{-2},T=5.24N$


Let $a$ be the common acceleration of the two masses $M_1$ and $M_2$ and $T$ be tension of the string.
Equation of motion for $M_1$ moving down $M_{1}gsin\theta +T-f_1=M_{1}a$

since $f_1={\mu }_1{M}_{1}gcos\theta$

$\therefore M_{1}gsin\theta -{\mu }_1{M}_{1}gcos\theta +T=M_{1}a\dots \dots \left(1\right)$

Equation of motion for $M_2$ moving down

$M_{2}gsin\theta -f_2-T=M_{2}a$

since $f_2={\mu }_2{M}_{2}gcos\theta$

$\therefore M_{2}gsin\theta -{\mu }_2{M}_{2}gcos\theta -T=M_{2}a\dots \dots \left(2\right)$

solving (1) and (2)

$a=\frac{(M_1+M_2)gsin\theta -gcos\theta ({\mu }_1{M}_1+{\mu }_2{M}_2)}{(M_1+M_2)}$

$=\frac{(4+2)\times 9.8\times 0.6-9.8\times 0.8(0.75\times 4+0.25\times 2)}{6}$

$=1.3m{s}^{-2}$

$T=M_{2}gsin\theta -{\mu }_2{M}_{2}gcos\theta -M_{2}a$

$=2\times 9.8\times 0.6-0.25\times 2\times 9.8\times 0.8-2\times 1.3$

$=5.24$ newton