Question

A force F⃗ of magnitude F making an angle θ with the x axis is applied to a particle located along axis of rotation A, at Cartesian coordinates (0,0) in the figure. The vector F⃗ lies in the xy plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the xy plane.

A particle is located at a vector position r⃗ r→r_vec with respect to an axis of rotation (thus r⃗ r→r_vec points from the axis to the point at which the particle is located). The magnitude of the torque ττtau about this axis due to a force F⃗ F→F_vec acting on the particle is given by

τ=rFsin(α)=(momentarm)⋅F=rF⊥τ=rFsin⁡(α)=(momentarm)⋅F=rF⊥,

where ααalpha is the angle between r⃗ r→r_vec and F⃗ F→F_vec, rrr is the magnitude of r⃗ r→r_vec, FFF is the magnitude of F⃗ F→F_vec, the component of r⃗ r→r_vec that is perpendicualr to F⃗ F→F_vec is the moment arm, and F⊥F⊥F_\perp is the component of the force that is perpendicular to r⃗ r→r_vec.

Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation.

In this problem, you must express the angle ααalpha in the above equation in terms of θθtheta, ϕϕphi, and/or ππpi when entering your answers. Keep in mind that π=180degrees and (π/2)=90degrees

A) What is the torque τA about axis A due to the force F_vec?.

B) What is the torque τB about axis B due to the force →F_vec? (B is the point at Cartesian coordinates (0,b)(0,b), located a distance b from the origin along the y axis.)

C) What is the torque τC about axis C due to →F_vec? (C is the point at Cartesian coordinates (c,0)(c,0), a distance c along the x axis.)

D) What is the torque τD about axis D due to →F_vec? (D is the point located at a distance d from the origin and making an angle ϕ with the x axis.)

Answer 1

BA(0,6) 90-0 Ē Ko A(0,0) @ 9=rXF ¢ (drosordein 17=11171 sing 0=60 19²0 Nm] (Ề = XP -1811Fising, = bfsin (90-0) y cobf coso (Clockwise direction) I = rfsino, = efsino Canticloerwise direction) | 8 =ffsing, --dfsin (6-0) (clockwise direction) 7xf will provide direction of torque.