Answer
a) $339.3 N \approx 340 N$
b) $1835.7 N \approx 1800 N$
Work Step by Step
The force required to produce the torque can be found from τ = rF sinθ . The force is applied
perpendicularly to the wrench, so θ = 90°.
$F=\frac{τ}{r}=\frac{95 m ⋅ N}{0.28 m}=339.3 N \approx 340 N$
The net torque still must be 95 m⋅N. This is produced by six forces, one at
each of the six points. We assume for our estimate that those forces are also perpendicular to their lever arms. From the diagram, we estimate the lever arm as follows, and then calculate the force at each point:
Lever arm $= r = \frac{1}{2}h+x=\frac{1}{2}(\frac{y}{cos 30^{\circ}})+y tan 30^{\circ}$
$=y(\frac{1}{2 cos 30^{\circ}}+tan 30^{\circ}) = (7.5\times10^{-3}m)(1.15)$
$τ_{net}=(6F_{point}r → F_{point}=\frac{τ}{6r}=\frac{95 m ⋅ N}{6(7.5\times10^{-3}m)(1.15)}=1835.7 N \approx 1800 N$