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.

Answer 1

a) τ = i ^ (y
`F_(z)` - z
`F_(y)`) + j ^ (z Fₓ - x
`F_(z)`) + k ^ (x
`F_(y)` - 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

Step-by-step explanation:

a) Torque is given by

τ = r x F

The easiest way to solve this equation is in the form of a determinant

`\tau =\left[\begin{array}{ccc}i&j&k\\x&y&z\\F_(x)&F_(y) &F_(z)\end{array}\right]`

The result is

τ = i ^ (y
`F_(z)` - z
`F_(y)`) + j ^ (z Fₓ - x
`F_(z)`) + k ^ (x
`F_(y)` - 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