A lower 0.56m long leg is being exercised. It has a 48.0N weight attached to the foot and is extended at an angle θ with respect to the vertical. Consider a rotation axis at the knee. At what angle θ does the magnitude of the torque equal 14.2 N-m?
To find the angle θ at which the magnitude of the torque equals 14.2 N-m, we can apply the principles of torque and rotational equilibrium. The torque is given by the equation:
τ = r * F * sin(θ),
where τ is the torque, r is the distance from the rotation axis to the point where the force is applied (in this case, the length of the leg), F is the force, and θ is the angle between the force vector and the lever arm.
In this case, the torque is given as 14.2 N-m, the length of the leg is 0.56 m, and the weight attached to the foot is 48.0 N.
Substituting these values into the torque equation, we have:
14.2 N-m = (0.56 m) * (48.0 N) * sin(θ).
To find the angle θ, we rearrange the equation:
sin(θ) = (14.2 N-m) / ((0.56 m) * (48.0 N)).
Now, we can solve for θ by taking the inverse sine of both sides:
θ = sin^(-1)((14.2 N-m) / ((0.56 m) * (48.0 N))).
Calculating this expression will give us the angle θ at which the magnitude of the torque equals 14.2 N-m.