Two ladders, 4.00 mand 3.00 mlong, are hinged at point Aand tied together by a horizontal rope 0.90 mabove the floor (Given figure). The ladders weigh 480 Nand 360 N, respectively, and the center of gravity of each is at its center. Assume that the floor is freshly waxed and frictionless.

1. Find the up- ward force at the bottom of each ladder.
2. Find the tension in the rope.
3. Find the magnitude of the force one ladder exerts on the other at point.
4. If an 800 Npainter stands at point, find the tension in the horizontal rope\

Here we have, the length of 1^st ladder is and the length of 2^nd ladder is

${\mathrm{L}}_2=3.0\mathrm{m}$

Weight of 1^st ladder is and weight of 2^nd ladder is

Height of rope is h = 0.9 m .

Which is says that,the object must also experience no net torque.

$\mathbf{\sum }_{\mathbf{}}\overrightarrow{\mathbf{\tau }}\mathbf{=}\mathbf{0}$

Where

$\mathit{\tau }$ is net external torque.

(a)

Consider the following free body diagram,

Now, normal forces balance the total weight must have

$w_1+w_2=N_1+N_2$

Now, by equation (1)

$\begin{array}{r}\sum \tau =0\\ N_1\left(L_1\cos \alpha +L_2\sin \alpha \right)-w_1\left[\left(\frac{L_1}{2}\right)\cos \alpha +L_2\sin \alpha \right]-w_2\left[\left(\frac{L_2}{2}\right)\sin \alpha \right]=0\end{array}$ (3)

Now, from the figure we can see that,

$\begin{array}{r}\tan \alpha =\frac{L_1}{L_2}\\ \tan \alpha =\frac{3.0\mathrm{m}}{4.0\mathrm{m}}\\ \alpha ={36.87}^{\circ }\end{array}$

Now substitute above value in equation (3), we get,

$\begin{array}{r}{N}_1\left(\left(4.0\mathrm{m}\right)\cos {36.87}^{\circ }+\left(3.0\mathrm{m}\right)\sin {36.87}^{\circ }\right)\\ -\left(480\mathrm{N}\right)\left[\left(\frac{4.0\mathrm{m}}{2}\right)\cos {36.87}^{\circ }+\left(3.0\mathrm{m}\right)\sin {36.87}^{\circ }\right]-\left(360\mathrm{N}\right)\left[\left(\frac{3.0\mathrm{m}}{2}\right)\sin {36.87}^{\circ }\right]=0\\ \left(5.0\mathrm{m}\right)N_1-1632\mathrm{N}\cdot \mathrm{m}-324.0\mathrm{N}\cdot \mathrm{m}=0\\ N_1=391\mathrm{N}\end{array}$

Now, from equation (2)

$\begin{array}{r}480\mathrm{N}+360\mathrm{N}=391\mathrm{N}+N_2\\ N_2=499\mathrm{N}\end{array}$

Hence, the up- ward force at the bottom of each ladder is${\mathrm{N}}_1=391\mathrm{N}\mathrm{and}{\mathrm{N}}_2=499\mathrm{N}$

(b)

From equation (1), we have, the net force in the horizontal direction has to be zero. Which means that tension forces acting at two sides are equal in magnitude.

The tension force can be found by considering the torque on the left ladder about the axis passing through the joint.

$\begin{array}{r}\begin{array}{rl}& \sum \tau =0\\ & N_1\cos \alpha L_1-W_1\cos \alpha \left(\frac{L_1}{2}\right)-T\left(L_1\sin \alpha -h\right)=0\end{array}\\ T=\frac{N_1\cos \alpha L_1-W_1\cos \alpha \left(\frac{L_1}{2}\right)}{\left(L_1\sin \alpha -h\right)}\end{array}$

By substituting the values in above equation. We get,

$\begin{array}{r}T=\frac{\left(391\mathrm{N}\right)\cos {36.87}^{\circ }\left(4.0\mathrm{m}\right)-\left(480\mathrm{N}\right)\cos {36.87}^{\circ }\left(\frac{\left(4.0\mathrm{m}\right)}{2}\right)}{\left(\left(4.0\mathrm{m}\right)\sin {36.87}^{\circ }-0.90\mathrm{m}\right)}\\ =322\mathrm{N}\end{array}$

Hence, tension in the rope is 322 N.

(c)

Both ladder exerts a force at point A on each other.

By, Newton’s third law that two force have equal magnitudes but in opposite direction.

In order to determine this force consider the right ladder. The horizontal and vertical forces exerted on it are

$F_{\text{horiz}}=T\text{and}{F}_{\text{vert}}=N_2-W_2$

Now,

$\begin{array}{r}{F}_A=\sqrt{{F_{\text{horiz}}}^2+{F_{\text{vert}}}^2}\\ =\sqrt{T^2+{\left(N_2-W_2\right)}^2}\\ =\sqrt{(322\mathrm{N}{)}^2+(449\mathrm{N}-360\mathrm{N}{)}^2}\\ =334\mathrm{N}\end{array}$

Hence, the magnitude of the force one ladder exerts on the other at point A is 334 N .

(d)

Let the painter weighing${\mathrm{w}}_{\mathrm{p}}=800\mathrm{N}$stands on point A.

Here the triangle which is made is not isosceles. So, weight won’t be distributed equally.

Let if 1^st ladder were vertical the force have to bear all of the additional weight.

To find the share of additional weight supported by each normal force, we have to look at the ratio of the horizontal distance from point to one end of the distance between two ends. We have,

$\begin{array}{r}\frac{L_2\sin \theta }{L_1\cos \theta +L_2\sin \theta }=\frac{\left(3.00\mathrm{m}\right)\sin {36.87}^{\circ }}{\left(4.00\mathrm{m}\right)\cos {36.87}^{\circ }+\left(3.00\mathrm{m}\right)\sin {36.87}^{\circ }}\\ =0.36\end{array}$

Which means that the force$N_1$will bear 36% of the painter’s weight.

Hence, the new forces are

$\begin{array}{r}{N}_1^{\mathrm{\prime }}=391\mathrm{N}+\left(0.36\right)w_p\\ =679\mathrm{N}\\ N_2^{\mathrm{\prime }}=499\mathrm{N}+\left(0.64\right)w_p\\ =961\mathrm{N}\end{array}$

Put above values in equation (3), we have,

$\begin{array}{r}T=\frac{\left(679\mathrm{N}\right)\cos {36.87}^{\circ }\left(4.0\mathrm{m}\right)-\left(480\mathrm{N}\right)\cos {36.87}^{\circ }\left(\frac{\left(4.0\mathrm{m}\right)}{2}\right)}{\left(\left(4.0\mathrm{m}\right)\sin {36.87}^{\circ }-0.90\mathrm{m}\right)}\\ =937\mathrm{N}\end{array}$

Hence, the tension in the horizontal rope If an 800 N painter stands at point is 937 N .