Question

a force of 4i-3j+5k is applied at a point whose position vector is 7i+4j_2k find the torque of force about the origin​

Answer 1

The torque is defined as the product of the force acting on the body and the distance which are both perpendicular to each other.

• The torque is written as τ=Nm, where N is the force acting on the body and the m is the unit of  distance between them.
• The given value of force F=4i-3j+5k.
• The distance is r= 7i+4j-2k.
• Now the dot product of force and distance is τ=(4i-3j+5k)·(7i+4j-2k)
• τ=(4·7+(-3·4)+(5·(-2))
• τ=28-12-10
• τ=28-22
• τ=6 Nm.
• As we know that i·i=j·j=k·k=1

Answer 2

The Torque of force about the origin is 6  Newton-meter

Explanation:

Given as :

The measure of force F = 4 i - 3 j + 5 k

The distance is  d = 7 i + 4 j - 2 k

Let The Torque of force about the origin = τ N-m

According to question

The torque is defined as the product of the force acting on the body and the distance which are both perpendicular to each other.

i.e  Torque = τ = F . d

where

N is the force acting on the body

d is the unit of distance between them

Now, We Apply dot products for vector

So, Torques = Force . distance

Or,  τ = ( 4 i - 3 j + 5 k ) · ( 7 i + 4 j - 2 k )

Or,  τ = [ ( 4 . 7 ) ( i . i ) + ( 4 . 4 ) ( i . j ) - ( 4 . 2 ) ( i . k ) ] + [ ( - 3 . 7 ) ( i . j ) + ( - 3 . 4 ) ( j . j ) + ( - 3 . - 2 ) ( j . k ) + ( 5 . 7 ) ( k . i ) ] + [ (5 . 7 ) ( k . i ) + ( 5 . 7) ( k . j ) + ( 5 . -2 ) ( k . k ) ]

Or, τ = [ 28 × 1 + 16 × 0 - 8 × 0 ] + [ - 21 × 0 - 12 × 1 + 6 × 0 ] + [  35 × 0 + 35 × 0 - 10  × 1 ]

∵ i · i = j · j = k · k = 1  and  i · j = j · k = k · i =  0

i.e τ = 28 + 0 - 0 - 0 - 12 + 0 + 0 + 0 - 10

Or, τ = 28 - 22

∴    Torque = 6  N -m

So, The Torque of force about the origin = τ = 6  N -m

Hence, The Torque of force about the origin is 6  Newton-meter . Answer