Question

) a wrench 0.6 meters long lies along the positive y-axis, and grips a bolt at the origin. a force is applied in the direction of ⟨0,1,3⟩ at the end of the wrench. find the magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt.

Answer 1

Applied Force Direction vector = [0,1,3] force F = j + 4k torque is t = 100 Newton-meters = r x F Wrench is 0.5 meters long on positive side of the y-axis, r = 0.6 = [0,0.6,0] We know torque |t| = |r x F| = |r| x |F| sin theta r x F = |r| x |F| cos theta r x (j + 3k) = |r| x |j + 3k| cos theta => [0,0.6,0] [0,1,3] = 0.6 x squareroot of ((0)^2 + (1)^2 + (3)^2) cos theta => 0.6 = 0.6 x squareroot of (1 + 9) cos theta => cos theta = 1 / squareroot of (10) Calculationg the sin theta, sin theta = squareroot of (1 - (1 / squareroot of (10))^2) = squareroot of (9/10) sin theta = 3 / squareroot of (10) Substituting the values, |T| = |r| x |F| sin theta => 100 = |0.6| x |F| x 3 / squareroot of (10) |F| = (100 x squareroot of (10)) / 1.8 |F| = (1000 / 18) x squareroot of (10) Magnitude of force |F| = 55.55 x squareroot of (10)