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

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) The torque can be expressed as

        τ = r x F

To tackle this equation, using the determinant approach is the most straightforward method

The resulting expression is

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

b) Now let's compute

     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) To find angular acceleration, we use

       τ = I α

       α = τ / I

The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.

     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