Ian: New page: Image:5.9.gif As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium. Resolving horisontally to the right $\begin{al...$

Ian — Tue, 06 Apr 2010 16:32:12 GMT

New page: Image:5.9.gif As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium. Resolving horisontally to the right <math>\begin{al...

New page

[[Image:5.9.gif]]

As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium.

Resolving horisontally to the right

<math>\begin{align}
& T2\cos {{28}^{\circ }}-T1\cos {{32}^{\circ }}=0 \\
& \\
\end{align}</math>

giving

<math>T2=\frac{\cos {{32}^{\circ }}}{\cos {{28}^{\circ }}}T1</math>

Resolving vertically upwards.

<math>T1\cos {{58}^{\circ }}+T2\cos {{62}^{\circ }}-mg=0</math>

Substituting for
<math>T2</math>
in this equation gives an equation only containing
<math>T1</math>.

<math>\begin{align}
& T1\cos {{58}^{\circ }}+\frac{\cos {{32}^{\circ }}}{\cos {{28}^{\circ }}}T1\cos {{62}^{\circ }}-mg=0 \\
& \\
& T1\left( \cos {{58}^{\circ }}+\frac{\cos {{32}^{\circ }}}{\cos {{28}^{\circ }}}\cos {{62}^{\circ }} \right)=mg \\
& \\
& T1\left( 0\textrm{.}53+0\textrm{.}45 \right)=40\times 9\textrm{.}8 \\
& \\
& T1=\frac{392}{0\textrm{.}98}=400\ \text{N} \\
\end{align}</math>

and thus using
<math>T2=\frac{\cos {{32}^{\circ }}}{\cos {{28}^{\circ }}}T1</math>

<math>T2=\frac{0\textrm{.}848}{0\textrm{.}883}\times 400=384\ \text{N}</math>

Solution 5.9 - Revision history

Ian: New page: Image:5.9.gif As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium. Resolving horisontally to the right \begin{al...

Ian: New page: Image:5.9.gif As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium. Resolving horisontally to the right $\begin{al...$