Question

Shafts A and B connect the gearbox to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x-axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and direction of its axis.

Answer is: M= 2860Nm

= 113.0 degrees

= 92.7 degrees

= 23.2 degrees

Answer 1

First find the unit vector along \(O A\). \(\lambda_{o_{1}}=-\sin 20 \mathrm{j}+\cos 20 \mathrm{k}\)

Now, take the moment about \(A\). Anti-clock wise moment is taken as positive. \(M_{A}=1600 \times(-\sin 20 \mathrm{j}+\cos 20 \mathrm{k})\)

\(=-547.23 \mathrm{j}+1503.5 \mathrm{k}\)

Now, find the unit vector along \(O B\). \(\lambda_{o B}=-\sin 20 \mathrm{j}-\cos 20 \mathrm{k}\)

Now, take the moment about \(B\).

$$ \begin{aligned} M_{B} &=-1200 \times(-\sin 20 \mathbf{j}-\cos 20 \mathrm{k}) \\ &=410.42 \mathrm{j}+1127.63 \mathrm{k} \end{aligned} $$

Now, take the moment about \(C\). \(M_{c}=-1120 \mathbf{i}\)

Find the resultant couple. We have, \(M=M_{A}+M_{3}+M_{C}\). \(M=-547.23 \mathrm{j}+1503.5 \mathrm{k}+410.42 \mathrm{j}+1127.63 \mathrm{k}-1120 \mathrm{i}\)

\(=-1120 \mathrm{i}-136.81 \mathrm{j}+2631.13 \mathrm{k}\)

Now, find the magnitude of the moment. \(M=\sqrt{(-1120)^{2}+(-136.81)^{2}+(2631.13)^{2}}\)

\(=2862.8 \mathrm{~N}\)

Now, \(\lambda=\frac{M}{|M|}\)

\(\lambda_{a d s}=\frac{\mathbf{M}}{M}\)

\(=\frac{-1120 \mathrm{i}-136.81 \mathrm{j}+2631.13 \mathrm{k}}{2862.7}\)

\(=-(0.3912) \mathbf{i}-(0.0477) \mathbf{j}+(0.919) \mathbf{k} 16\)

Now, find the angles. \(\cos \left(\theta_{x}\right)=-0.3912\)

\(\theta_{x}=\cos ^{-1}(-0.3912)\)

\(=113^{\circ}\)

\(\cos \left(\theta_{y}\right)=0.0477\)

\(\theta_{y}=\cos ^{-1} 0.0477\)

\(=92.7^{\circ}\)

\(\cos \left(\theta_{z}\right)=0.919\)

\(\theta_{z}=\cos ^{-1} 0.919\)

\(=23.2^{\circ}\)