Torque Revisited — Question 9

Question 9

A spatial cable creates more than the desired axle moment

A cable attaches at $P=(2,0,0)\,\mathrm m$ and runs toward anchor $B=(0,3,6)\,\mathrm m$. Its tension is $T$. The target torque about the $z$ axis through the origin is $12\,\mathrm{N\,m}$.

1. Find required tension.

2. Find the full torque and identify the transverse component that the support must handle.

Question 9 - Solution

A spatial cable creates more than the desired axle moment

(a) Derivation and calculation

The cable direction is $$\overrightarrow{PB}=(-2,3,6)\ \mathrm m,\qquad|PB|=\sqrt{4+9+36}=7\ \mathrm m.$$ Thus $\vec F=T(-2,3,6)/7$ and $$\vec\tau=(2,0,0)\times\vec F=(0,-12T/7,6T/7).$$ Set its axial component to the target: $$6T/7=12\Rightarrow\boxed{T=14.0\ \mathrm N}.$$

(b) Result and interpretation

The resulting force and torque are $$\vec F=(-4,6,12)\ \mathrm N,\qquad \boxed{\vec\tau=(0,-24,12)\ \mathrm{N\,m}}.$$ The transverse part $\boxed{-24\hat y\ \mathrm{N\,m}}$ must enter the support/bearing moment balance. A scalar $z$-axis calculation does not account for it.

Physical check

The axial torque is $r_xF_y=2(6)=12$ N m, while the vertical cable force creates the separate $y$ moment.