A rod of length 20cm is kept along x axis. Two forces 5N and 10N are applied at distances 10cm and 15cm from point A as shown in figure. Find the point of application of net force from point A.
A
20cm
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B
12.5cm
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C
15cm
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D
17.5cm
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Solution
The correct option is A20cm Given, length of rod (l) = 20cm
Position vector of point of application of F1=5N(−^j) ¯r1=10cm^i
Position vector of point of application of F2=10N(^j) ¯r2=15cm^i
Torque due to F1 →τ1=→r1×→F1=50×10−2Nm(−^k)
Torque due to F2 →τ2=r2×→F2=150×10−2Nm(^k)
Net force acting on rod (Fnet)=→F1+→F2=5N(^j)
Since, →r1×→F1+→r2×→F2=→r×→Fnet we get 150×10−2−50×10−2=5r⇒r=20cm