To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to

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1 year ago

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To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to

rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam.
Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r.
Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 2.19m, ry = 0.025 m, rz = 0.035 m, Fx = -7.8 N, Fy = 1.2 N, Fz = 2.2 N.
Part (c) If the moment of inertia of the beam with respect to the pivot is I = 494 kgË™m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared.

Physics

2 answers:

Llana [10] · Answer 1 · 2024-08-11T23:51:01+03:00

Answer:

1. Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

2. Magnitude of resulting torque = 3.84 Nm

3. Angular acceleration = = 0.0086877 rad/s² rad/s²

Explanation:

It is given that:

I = 442 kg˙m2

rx = 0.76 m, ry = 0.035 m, rz = 0.015 m,

Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N

F = Fx i + Fy j + Fz (equation 1)

r = rx i + ry j + rz k (equation 2)

(a) Torque

T = r * F (equation 3)

by putting equation 1 and 2 into equation 3, we have;

Torque= r x F

= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )

= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

Therefore,

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

b) Magnitude of the torque

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

=(0.025 x 2.2 - (0.035 x (-1.2))) i +(0.035 x (-7.8)) - 2.19 x 2.2) j+(2.19 x 1.2 - 0.025 - (-7.8)) k

= (0.055 - 0.042) i + (0.273 - 4.818) j + (-2.628 -0.195) k

= (0.013) i - (-4.545) j + (-2.433) k

Magnitude of resulting torque = √0.013² + 4.545² - 2.433²

=√14.751

= 3.84 Nm

c) Angular acceleration

Since, angular acceleration = torque/moment of inertia

= 3.84/ 442

= 0.0086877 rad/s²

Hence,

1. Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

2. Magnitude of resulting torque = 3.84 Nm

3. Angular acceleration = = 0.0086877 rad/s² rad/s²

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natali 33 [55] · Answer 2 · 2024-08-11T21:29:49+03:00

Answer:

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Magnitude of resulting torque = 3.84 Nm
Angular acceleration = = 0.0086877 rad/s² rad/s²

Explanation:

It is given that:

I = 442 kg˙m2

rx = 0.76 m, ry = 0.035 m, rz = 0.015 m,

Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N

F = Fx i + Fy j + Fz (equation 1)

r = rx i + ry j + rz k (equation 2)

(a) Torque

T = r * F (equation 3)

by putting equation 1 and 2 into equation 3, we have;

Torque= r x F

= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )

= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

Therefore,

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

b) Magnitude of the torque

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

=(0.025 x 2.2 - (0.035 x (-1.2))) i +(0.035 x (-7.8)) - 2.19 x 2.2) j+(2.19 x 1.2 - 0.025 - (-7.8)) k

= (0.055 - 0.042) i + (0.273 - 4.818) j + (-2.628 -0.195) k

= (0.013) i - (-4.545) j + (-2.433) k

Magnitude of resulting torque = √0.013² + 4.545² - 2.433²

=√14.751

= 3.84 Nm

c) Angular acceleration

Since

angular acceleration = torque/moment of inertia

= 3.84/ 442

= 0.0086877 rad/s²

Hence,

Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Magnitude of resulting torque = 3.84 Nm
Angular acceleration = = 0.0086877 rad/s² rad/s²

To learn more about Torque, use the given link:

brainly.com/question/19247046?referrer=searchResults

#SPJ4