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Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?

Respuesta :

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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