Answer:
Torque, [tex]\tau=0i+0j-70k[/tex]
Explanation:
It is given that,
Force acting on the particle, [tex]F=-10j\ N[/tex]
Position of the particle, [tex]r=(7i+5j)\ m[/tex]
We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :
[tex]\tau=r\times F[/tex]
[tex]\tau=(7i+5j)\times (-10j)[/tex]
The cross product of vectors is given by :
[tex]\tau=\begin{pmatrix}0&0&-70\end{pmatrix}[/tex]
or
[tex]\tau=0i+0j-70k[/tex]
So, the torque on the particle about the origin [tex]0i+0j-70k[/tex]. Hence, this is the required solution.
The torque on the particle about the origin is [tex]0i-0j-70kNm[/tex]
The formula for calculating the torque is expressed as:
[tex]\tau = F \times r[/tex]
F is the force applied
r is the radius of the particle;
Given the following parameters
[tex]F=-10j N\\r=(7i+5j)m[/tex]
Substitute the given parameter into the formula:
[tex]\tau=-10j \times (7i+5j)[/tex]
According to cross product, [tex]i \times j = k, j \times i=-k[/tex]
Taking the cross product of the force and the radius is [tex]0i-0j-70k[/tex]
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