Question

Two blocks connected by a massless string slides down 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 (a) the common acceleration of two masses and (b) the tension in the string. $(\sin 37^{\circ }=0.6,\cos 37^{\circ }=0.8)$.(Take $g=9.8m/s^2)$

Answer 1

Maximum force of friction between $M_1$ and inclined plane
$f_1=\mu M_{1}g\cos \theta =(0.75)(4)(9.8)(0.8)=23.52N$
$M_{1}g\sin \theta =(4)(9.8)(0.6)=23.52N=F_1$ (say)
Maximum force of friction between $M_2$ and inclined plane
$f_2={\mu }_2{M}_{2}g\cos \theta$
$=(0.25)(2)(9.8)(0.8)=3.92N$
$M_{2}g\sin \theta =(2)(9.8)(0.6)=11.76N=F_2$ (say)

Both the blocks will be moving downwards with same acceleration $a$. Different forces acting on two blocks are as shown in above figure.
Equation of motion of $M_1$
$T+F_1-f_1=M_{1}a$
or $T=4a...(i)$
Equation of motion of $M_2$
$F_2-T-f_2=M_{2}a$
or $7.84-T=2a...(i)$
Solving Eqs. (i) and (ii), we get
$a=1.3m/s^2$ and $T=5.2N$