Question

Find the torque of a force F=-3I+2J+K acting at the point r=8i+2j+3k

Answer 1

Answer:

-4i-17j+22k

Explanation:

Now,torque is the cross product of position vector (r) and force (F) ; denoted by (r×F).torque is a vector quantity.

Here,Torque,=(2*1-3*2)i +{3*(-3)-8*1}j + {8*2-2*(-3)}k

= -4i-17j+22k

Answer 2

Concept Introduction:-

Torque is the in physics as well as mechanics, the linear force's equivalent.

Given Information:-

We have been given that a  force $F=-3I+2J+K$ acting at the point $r=8i+2j+3k$

To Find:-

We have to find that the torque of a force.

Solution:-

According to the problem

$r = 8i + 2j + 3k$ and

$F = -3i + 2j + k$

Now, torque is the cross product of position vector (r) and force (F); denoted by $(r \times F)$. Torque is a vector quantity.

Here, Torque$= (2\times 1 - 3\times 2)i + {3\times (-3) - 8\times 1}j + {8\times 2 - 2\times (-3)}k= -4i -17j + 22k$.

This torque vector will operate through the origin $(0, 0, 0)$; perpendicular to both r and F, in the appropriate direction as obtained from above calculation.

Hence, magnitude of this torque vector$= \sqrt{[{(-4)^2} + {(-17)^2} + (22^2)] }$force units

$=\sqrt{(16 + 289 + 484)}$ force units

$=\sqrt{789}$ force units

≈$28.09$ force units.

Final Answer:-

The torque is $28.09$ force units.

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