CBSE-XI-Physics
05: Newton's Laws of Motion
- #28Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5-E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?
Figure
Ans : The free-body diagram for mass m1 is shown below:
(Figure 1)
The free-body diagram for mass m2 is shown below:
(Figure 2)
The free-body diagram for mass m3 is shown below:
(Figure 3)
Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.
So, the actual acceleration of the blocks m1, m2 and m3 will be a1, (a1 - a2) and (a1 + a2), as shown.
From figure 2, T - 1g - 1a1 = 0 ...(i)
From figure 3, `` \frac{T}{2}-2g-2\left({a}_{1}-{a}_{2}\right)=0...\left(\,\mathrm{\,ii \,}\right)``
From figure 4, `` \frac{T}{2}-3g-3\left({a}_{1}+{a}_{2}\right)=0...\left(\,\mathrm{\,iii \,}\right)``
From equations (i) and (ii), eliminating T, we get:
1g + 1a2 = 4g + 4 (a1 + a2)
5a2 - 4a1 = 3g ...(iv)
From equations (ii) and (iii), we get:
2g + 2(a1 - a2) = 3g - 3 (a1 - a2)
5a1 + a2 = g ...(v)
Solving equations (iv) and (v), we get:
`` {a}_{1}=\frac{2g}{29}``
`` \,\mathrm{\,and \,}{a}_{2}=g-5{a}_{1}``
`` \Rightarrow {a}_{2}=g-\frac{10g}{29}=\frac{19g}{29}``
`` \,\mathrm{\,So \,},{a}_{1}-{a}_{2}=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29}``
`` \,\mathrm{\,and \,}{a}_{1}+{a}_{2}=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29}``
So, accelerations of m1, m2 and m3 are `` \frac{19g}{29}\left(\,\mathrm{\,up \,}\right),\frac{17g}{29}\left(\,\mathrm{\,down \,}\right)\,\mathrm{\,and \,}\frac{21g}{29}\left(\,\mathrm{\,down \,}\right)`` , respectively.
Now, u = 0, s = 20 cm = 0.2 m
`` {a}_{2}=\frac{19g}{29}``
`` \therefore s=ut+\frac{1}{2}a{t}^{2}``
`` \Rightarrow 0.2=\frac{1}{2}\times \frac{19}{29}g{t}^{2}``
`` \Rightarrow t=0.25\,\mathrm{\,s \,}``
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