# -what is the magnitude of tension in the rope? -find the horizontal and vertical components of the force exerted at A on the lamina by the wall.

## A uniform triangular lamina ABC is right-angled at B and has sides AB = 0.6 m and BC = 0.8 m. The mass of the lamina is 4 kg. One end of a light inextensible rope is attached to the lamina at C. The other end of the rope is attached to a fixed point D on a vertical wall. The lamina is in equilibrium with A in contact with the wall at a point vertically below D. The lamina is in a vertical plane perpendicular to the wall, and AB is horizontal. The rope is taut and at right angles to AC Find (i) the tension in the rope, (ii) the horizontal and vertical components of the force exerted at A on the lamina by the wall.

Mar 3, 2018

See below.

#### Explanation:

For a homogeneous lamina, the triangle's center of mass is given by

$G = \frac{1}{3} \left(A + B + C\right)$

Now momentum equilibrium

${\vec{M}}_{A} = \left(C - A\right) \times \vec{V} + \left(G - A\right) \times \vec{P} = \vec{0}$

and static resultant equilibrium

$\vec{R} = \vec{V} + \vec{H} + \vec{P} = \vec{0}$

Here

$A = \left(0 , 0\right)$
$B = \left(0.6 , 0\right)$
$C = \left(0.6 , 0.8\right)$
$G = \left(0.4 , 0.26667\right)$ Lamina center of mass.
$\vec{V} = v \left(- \cos {\theta}_{0} , \sin {\theta}_{0}\right)$ Rope tension force.
$\vec{H} = \left({h}_{x} , {h}_{y}\right)$ Reaction force at point $A$.
$\vec{P} = m g \left(0 , - 1\right)$ Lamina weight .

${\theta}_{0} = \arctan \left(\frac{\overline{C B}}{\overline{A B}}\right)$

After substituting values on ${\vec{M}}_{A} = \vec{0}$ and $\vec{R} = \vec{0}$ we obtain

$\left\{\begin{matrix}0.96 v - 16 = 0 \\ {h}_{x} - 0.6 v = 0 \\ {h}_{y} + 0.8 v = 40\end{matrix}\right.$

giving

$v = 16.6667$
${h}_{x} = 10$
${h}_{y} = 26.6667$

or

$\vec{V} = 11.6667 \left(- \cos {\theta}_{0} , \sin {\theta}_{0}\right)$
$\vec{H} = \left(10 , 26.6667\right)$