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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 freel
Question
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 = 4.07 m, ry = 0.075 m, rz = 0.035 m, Fx = 2.8 N, Fy = 8.4 N, Fz = 1.4 N. Part (c) If the moment of inertia of the beam with respect to the pivot is I = 241 kg˙m², calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared.
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Physics
4 years
2021-08-18T22:23:51+00:00
2021-08-18T22:23:51+00:00 1 Answers
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Answers ( )
Answer:
a) τ = i ^ (y
– z 
) + j ^ (z Fₓ – x 
) + k ^ (x 
– y Fₓ)
b)τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c)α = (-7.8 10⁻⁴ i ^ – 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Explanation:
a) Torque is given by
τ = r x F
The easiest way to solve this equation is in the form of a determinant
The result is
τ = i ^ (y
– z 
) + j ^ (z Fₓ – x 
) + k ^ (x 
– y Fₓ)
b) let’s calculate
τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 – 4.07 1.4) + k ^ (4.07 8.4 – 0.075 2.8)
τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)
τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) calculate the angular acceleration with
τ = I α
α = τ / I
Since the moment of inertia is a scalar, the direction does not change, only the modulus of each element changes.
α = (-0.189i^ -5.6 j^ + 33.978k^) / 241
α = (-7.8 10⁻⁴ i ^ – 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²