Multiplying Vectors — Question 8

Question 8

Work and torque use different products

A constant force is $\vec F=(3,4,0)$ N and a displacement is $\vec d=(2,-1,0)$ m. Define work by $W=\vec F\cdot\vec d$. Independently, the force is applied at position $\vec r=(2,-1,0)$ m relative to an origin; define torque by $\vec\tau=\vec r\times\vec F$. No dynamical equations are needed.

1. Find W and the full torque vector with units.

2. Do equal dimensional units make work and torque interchangeable? Explain their different directional meanings.

3. Shift the torque reference origin to the point of application while keeping the force and displacement fixed. Find the new torque and work.

Question 8 - Solution

Work and torque use different products

(a) Apply the specified order and product.

$$\boxed{W=3(2)+4(-1)=2\,\mathrm J},\qquad\vec\tau=[2(4)-(-1)(3)]\hat k=11\hat k\,\mathrm{N\,m}$$ $W=3(2)+4(-1)=2\ \mathrm{N\,m}=2$ J. The torque is $\vec\tau=[2(4)-(-1)(3)]\hat k=11\hat k\ \mathrm{N\,m}$. It points out of the xy plane by the right-hand rule.

(b) Retain type and physical definition.

$$W=Fd\cos\theta\quad\text{(scalar)},\qquad\tau=rF\sin\phi\quad\text{(magnitude of axial vector)}$$ Work is a scalar measuring the force component along displacement times displacement length. Torque is an axial vector describing the moment of force about the chosen origin. Both have dimensions force times length, but they are not the same quantity. Torque is conventionally written in N m rather than joules to preserve that distinction.

(c) Change the lever arm, not the displacement.

$$\vec r'=\vec0\Rightarrow\vec\tau'=\vec0\times\vec F=\vec0,\qquad W'=\vec F\cdot\vec d=2\,\mathrm J$$ At the new origin $\vec r\,{}^{\prime}=\vec0$, so $\vec\tau\,{}^{\prime}=\vec0$. Work stays $\vec F\cdot\vec d=2$ J because the specified displacement and force have not changed. Torque generally depends on the reference origin; the work of this constant force over a fixed displacement does not.