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Question: Find the torque of a force \(7\hat i + 3\hat j - 5\hat k\) about the origin. The force acts on a par...

Find the torque of a force 7i^+3j^−5k^7\hat i + 3\hat j - 5\hat k about the origin. The force acts on a particle whose position vector is i^−j^+k^\hat i - \hat j + \hat k.

Explanation

Solution

In this question, discuss the significance of torque in motion of the body. Then discuss the formula of torque. The torque is the cross product of the position vector and the vector form of the force applied.

Complete step by step solution:
We know that the force is what causes a body to accelerate in linear kinematics. Similarly, we know that torque is the tendency of a force to cause or change the rotational motion of a body. In another word it is also called the twist or turning force on a body. Generally, we calculate torque of any body by multiplying the force with displacement. It is a vector quantity. It has both direction and magnitude.
We know that the torque is calculated as,
τ=r×F\tau = r \times F
Where, τ\tau is the torque applied on an object
rr is the position vector
FF is the applied Force vector.
Here, it is given that the applied force is F=7i^+3j^−5k^F = 7\hat i + 3\hat j - 5\hat k and the given position vector is r=i^−j^+k^r = \hat i - \hat j + \hat k.
Now, we substitute the given values in the torque equation as,
τ=r×F\tau = r \times F
(i^−j^+k^)×(7i^+3j^−5k^)(\hat i - \hat j + \hat k) \times (7\hat i + 3\hat j - 5\hat k)
We will use the determinant rule for solving this cross product.
\tau = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&{ - 1}&1 \\\ 7&3&{ - 5} \end{array}} \right|
τ=(5−3)i^−(−5−7)j^+(3+7)k^\tau = (5 - 3)\hat i - ( - 5 - 7)\hat j + (3 + 7)\hat k
On simplification we get,
τ=2i^+12j^+10k^\tau = 2\hat i + 12\hat j + 10\hat k
Therefore, the torque of a force 7i^+3j^−5k^7\hat i + 3\hat j - 5\hat k about the origin is τ=2i^+12j^+10k^\tau = 2\hat i + 12\hat j + 10\hat k.

Note: During this calculation we must be very careful during the vector cross product. For determining the direction of torque, we will follow the right-hand rule. With your right-hand, point your index finger along the vector rr, and point your middle finger along the vector FF, then the cross product that is the direction of torque goes in the direction of your thumb.