O 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. show answer No Attempt 33% 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. τ = | γ θ i j k d Fx Fy Fz g m n rx ry rz ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 0.76 m, ry = 0.035 m, rz = 0.015 m, Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N. No Attempt No Attempt 33% Part (c) If the moment of inertia of the beam with respect to the pivot is I = 442 kg˙m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared. All content © 2020 Expert TA, LLC

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

(b) Magnitude of resulting torque  =  = 3.99 Nm

(c) angular acceleration = = 0.009027 rad/s²

Explanation:

Given Data;

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 1

r =  rx i + ry j + rz k 2

(a) Torgue is given by the formula;

T = r * F  3

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)

Putting given values into the above expression, we have

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

=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k

= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k

= (0.196) i - (3.29) j + (-2.254) k

Magnitude of resulting torque = √(0.196² + 3.29² +2.254²

                                                  =√15.943031

                                                  = 3.99 Nm

c) Angular acceleration is given by the formula;

angular acceleration = torque/moment of inertia

                                   = 3.99/ 442

                                  = 0.009027 rad/s²

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray