A two dimensional object placed in the xy plane has three forces acting on it: a force of 3N along the x axis acting at a point (3m, 4m); A force of 2N along the y axis acting at (-2, 5m); and a force of 5N in the negative x direction acting at (-2m, -3m). What is the net torque about the point (-1m, 1m)?
To find the net torque about a point, we need to calculate the torque contributed by each individual force and then sum them up.
The formula for torque is given by the cross product of the vector from the point of rotation to the point where the force is applied and the force vector itself. Mathematically, the torque vector (τ) can be calculated as τ = r x F, where 'x' denotes the cross product.
Let's calculate the net torque contributed by each force.
1) Force of 3N along the x-axis at (3m, 4m):
The vector from the point of rotation (-1m, 1m) to the point where the force is applied (3m, 4m) is given by r = (3m, 4m) - (-1m, 1m) = (4m, 3m).
The force vector is F = (3N, 0).
Calculating the cross product r x F gives us: τ1 = (4m x 0) - (3m x 3N) = 0 - 9Nm = -9Nm.
2) Force of 2N along the y-axis at (-2m, 5m):
The vector from the point of rotation (-1m, 1m) to the point where the force is applied (-2m, 5m) is given by r = (-2m, 5m) - (-1m, 1m) = (-1m, 4m).
The force vector is F = (0, 2N).
Calculating the cross product r x F gives us: τ2 = (-1m x 2N) - (4m x 0) = -2Nm - 0 = -2Nm.
3) Force of 5N in the negative x-direction at (-2m, -3m):
The vector from the point of rotation (-1m, 1m) to the point where the force is applied (-2m, -3m) is given by r = (-2m, -3m) - (-1m, 1m) = (-1m, -4m).
The force vector is F = (-5N, 0).
Calculating the cross product r x F gives us: τ3 = (-1m x 0) - (-4m x -5N) = 0 - 20Nm = -20Nm.
Now, we can calculate the net torque (τ_net) by summing up the individual torques:
τ_net = τ1 + τ2 + τ3
= -9Nm + (-2Nm) + (-20Nm)
= -31Nm
Therefore, the net torque about the point (-1m, 1m) is -31Nm.