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A force F~ = (1.8 N)ˆı + (3 N) ˆ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at the point R~ = (4.1 m)ˆı + (5.4 m) ˆ. Find the z-component of the net torque. Answer in units of N m.

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joc0916

Answer:

The z-component of the net torque is: 2.58 (N*m)

Explanation:

We need to apply the torque equation ([tex]T=RXF[/tex]), where T is the torque, R is the distance at the point where the force is applied and F is the force, and remembering the cross product ([tex]AXB=abSin(\beta)(n)[/tex]), where A and B are the vectors and ab are the magnitudes, (beta) the angle between A and B in the plane containing them, and (n) is the direction of the vector given by the right-hand rule. Knowing those things, we can get:[tex]T=RXF=4.1(i)+5.4(j)X1.8(i)+3(j)=0+4.1*3(k)+5.4*1.8(-k)+0=2.58(k)(N*m)[/tex] that is the z-component of the torque.

The torque on the object is -2.58Nm

Data;

  • F = (1.8i + 3j)N
  • r = (4.1i + 5.4j)m

Net Torque

The net torque is calculated as the product of the force and distance

[tex]t = r * F[/tex]

Let's substitute the values and solve

Since this is a dot product, we can easily multiply through

[tex]\tau = r * F\\\tau = (1.8*5.4) - (3 * 4.1) = -2.58Nm[/tex]

Using 2 by 2 matrix method, the torque on the object is -2.58Nm

Learn more on torque here;

https://brainly.com/question/20691242

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