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Exercise 6.1 · Q13

Q.Find the magnitude and direction cosines of the torque of a force represented by 3i^+4j^−5k^3\hat i+4\hat j-5\hat k about the point with position vector 2i^−3j^+4k^2\hat i-3\hat j+4\hat k acting through a point whose position vector is 4i^+2j^−3k^4\hat i+2\hat j-3\hat k.

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Torque about a point equals the cross product of (position vector of the force's point of application, RELATIVE to the moment point) with the force itself; its magnitude and direction cosines follow from the resulting vector.

Step 1. Find r⃗\vec r relative to the moment point. The force acts through the point with position vector 4i^+2j^−3k^4\hat i+2\hat j-3\hat k; torque is taken about 2i^−3j^+4k^2\hat i-3\hat j+4\hat k. So

r⃗=(4−2)i^+(2−(−3))j^+(−3−4)k^=2i^+5j^−7k^.\vec r=(4-2)\hat i+(2-(-3))\hat j+(-3-4)\hat k=2\hat i+5\hat j-7\hat k.

Step 2. Compute the torque τ⃗=r⃗×F⃗\vec\tau=\vec r\times\vec F, with F⃗=3i^+4j^−5k^\vec F=3\hat i+4\hat j-5\hat k:

τ⃗=∣i^j^k^25−734−5∣=i^(5(−5)−(−7)(4))−j^(2(−5)−(−7)(3))+k^(2(4)−5(3)).\vec\tau=\begin{vmatrix}\hat i&\hat j&\hat k\\ 2&5&-7\\ 3&4&-5\end{vmatrix}=\hat i\big(5(-5)-(-7)(4)\big)-\hat j\big(2(-5)-(-7)(3)\big)+\hat k\big(2(4)-5(3)\big).

=i^(−25+28)−j^(−10+21)+k^(8−15)=3i^−11j^−7k^.=\hat i(-25+28)-\hat j(-10+21)+\hat k(8-15)=3\hat i-11\hat j-7\hat k. …

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