Laws of Motion 3 Question 23
24. 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 $M _1=4 kg$ and $M _2=2 kg$ 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. $\left(\sin 37^{\circ}=0.6, \cos 37^{\circ}=0.8\right) .\left(\right.$ Take $\left.g=9.8 m / s^{2}\right)$
(1979)
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Answer:
Correct Answer: 24. (a) $a=1.3 m / s^{2}$
(b) $T=5.2 N$
Solution:
- Maximum force of friction between $M _1$ and inclined plane
$$ \begin{aligned} f _1 & =\mu _1 M _1 g \cos \theta=(0.75)(4)(9.8)(0.8)=23.52 N \\ M _1 g \sin \theta & =(4)(9.8)(0.6)=23.52 N=F _1 \end{aligned} $$
Maximum force of friction between $M _2$ and inclined plane
$$ \begin{aligned} f _2 & =\mu _2 M _2 g \cos \theta \\ & =(0.25)(2)(9.8)(0.8)=3.92 N \\ M _2 g \sin \theta & =(2)(9.8)(0.6)=11.76 N=F _2 \end{aligned} $$
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 $M _1$
$$ \text { or } \quad \begin{aligned} T+F _1-f _1 & =M _1 a \\ T & =4 a \end{aligned} $$
Equation of motion of $M _2$
$$ \begin{aligned} F _2-T-f _2 & =M _2 a \\ \text { or } \quad 7.84-T & =2 a \end{aligned} $$
Solving Eqs. (i) and (ii), we get
$m g \sin \theta=(2)(10) \frac{1}{2}=10 N=F _1$
(a) Force required to move the block down the plane with constant velocity.
$F _1$ will be acting downwards, while $F _2$ upwards.
Since $F _2>F _1$, force required
$$ F=F _2-F _1=11.21 N $$
(b) Force required to move the block up the plane with constant velocity.
$F _1$ and $F _2$ both will be acting downwards.
$$ \therefore \quad F=F _1+F _2=31.21 N $$
