Two block of mass mA=10kg and mB=15kg are connected by a light inextensible string and placed on a smooth horizontal surface as shown in the figure. Find the value of tension force on block A, if force applied on block B is 50√3N .
A
30N
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B
45N
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C
15N
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D
75N
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Solution
The correct option is A30N ⇒The forces in vertical direction are balanced for both bodies, because system will be moving only in horizontal direction.
From the FBD of blocks:
For block A T=mAa=10a...(i)
For block B Fcosθ−T=mBa=15a...(ii)
Adding equation (i) and (ii): Fcosθ=25a 75=25a ∴a=7525=3m/s2 ⇒T=mAa=10×3=30N