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  • Two blocks of mass 5kg and 5kg are connected by a string of mass 2kg as shown in fig. Point Q is the midpoint of the string. A pulling force of 24N is applied on block 5kg at an angle of <img

Two blocks of mass 5kg and 5kg are connected by a string of mass 2kg as shown in fig. Point Q is the midpoint of the string. A pulling force of 24N is applied on block 5kg at an angle of 60^{\circ} with horizontal. The tension (in N) at point Q is (all surfaces are frictionless).

Option: 1

6


Option: 2

10


Option: 3

5


Option: 4

10


Answers (1)

best_answer

Solution :

\\ \text{Let a be the horizontal acceleration of the whole system due to the horizontal component of force } F\cos60^{\circ}

\therefore a=\frac{\text{Net force along horizontal direction}}{\text{Total mass}} = \frac{24\cos60^{\circ}}{5+5} =\frac{12}{10}=\frac{5}{6} \ m/s^{2}

\\ \text{FBD of the 5 Kg block :}

\\ \text{Using Newton's second law, } \\ T=ma=5\times \frac{6}{5}=6 N

 

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