Question:
A triangular plate is shown. A force \(\vec{F}=4\hat{i}-3\hat{j}\) is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :
A triangular plate is shown. A force \(\vec{F}=4\hat{i}-3\hat{j}\) is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :
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Find the position of P relative to O and to Q (equilateral triangle of side 10 cm), then take the z-component of r x F.
Updated On: Nov 4, 2026
- \(-15+20\sqrt{3},\ 15+20\sqrt{3}\)
- \(15-20\sqrt{3},\ 15+20\sqrt{3}\)
- \(15+20\sqrt{3},\ 15-20\sqrt{3}\)
- \(-15-20\sqrt{3},\ 15-20\sqrt{3}\)
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The Correct Option is D
Solution and Explanation
The triangle is equilateral with side 10 cm. Taking O as origin: Q = (10, 0) cm and P = (5, \(5\sqrt{3}\)) cm.
For a planar vector, \(\tau_z = xF_y - yF_x\) with \(F_x=4,\ F_y=-3\).
About O: \(\vec r=(5,\,5\sqrt3)\), \(\tau_O=5(-3)-5\sqrt3(4)=-15-20\sqrt3\).
About Q: \(\vec r=\vec{QP}=(-5,\,5\sqrt3)\), \(\tau_Q=(-5)(-3)-5\sqrt3(4)=15-20\sqrt3\).
(Values are in N cm for r in cm.)
Answer: (4) \(-15-20\sqrt{3},\ 15-20\sqrt{3}\)
For a planar vector, \(\tau_z = xF_y - yF_x\) with \(F_x=4,\ F_y=-3\).
About O: \(\vec r=(5,\,5\sqrt3)\), \(\tau_O=5(-3)-5\sqrt3(4)=-15-20\sqrt3\).
About Q: \(\vec r=\vec{QP}=(-5,\,5\sqrt3)\), \(\tau_Q=(-5)(-3)-5\sqrt3(4)=15-20\sqrt3\).
(Values are in N cm for r in cm.)
Answer: (4) \(-15-20\sqrt{3},\ 15-20\sqrt{3}\)
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