Question

A 10.0 kg uniform ladder that is 2.50 m long is placed against a smooth vertical wall and reaches to a height of 2.10 m, as shown in the figure. The base of the ladder rests on a rough horizontal floor whose coefficient of static friction with the ladder is 0.800. An 80.0 kg bucket of concrete is suspended from the top rung of the ladder, right next to the wall, as shown in the figure. 

a) What is the magnitude of the force that the floor exerts on the ladder?

b) What is the magnitude of the friction force that the wall exerts on the ladder?

c) What is the magnitude of the friction force that the floor exerts on the ladder?

Answer 1

a. magnitude of force exerrted by floor = sqrt(538^2 + ((80+10)*9.8)^2) = 1033 N

b. Friction force by wall = 0 as it is smooth.

c. Distance of lower end from wall d = sqrt(2.5^2-2.1^2) = 1.356 m

Let normal by wall be N1 and N2 by ground,

balancing torque about lower end,

mg d/2 + Mgd = N1*2.1

N1 = [1.356*10*9.8/2 + 1.356*80*9.8]/2.1

= 538 N

N1 will be balanced by friction by ground,

Friction by floor = 538N

Answer 2

The normal reaction is equal to sum of the weights, so the maximum frictional force is

$$ \begin{aligned} F_{f-\max } &=\mu F_{N} \\ &=0.8 \times 9.8 \times(10+80) \\ &=705.6 \mathrm{~N} \end{aligned} $$

When the mass is at2.5 \(\mathrm{m}\), the normal force in the \(\mathrm{x}\) directioncounters the torque due to the net

mass of the mass and the ladder. The normal force due to wall is thus

$$ \begin{aligned} F_{N-\text { val }}=& \stackrel{\vec{F}_{\text {mass }} \times \vec{r}_{\text {mas }}+\vec{F}_{\text {ladler }} \times\left(\frac{\vec{r}_{\text {ladder }}}{2}\right)}{\vec{r}_{\text {ladder }} \sin \left(90^{\circ}-\theta\right)} \\ &=\frac{9.8 \times \sin \theta\left(80 \times 2.5+10 \times \frac{2.5}{2}\right)}{2.5 \times \cos \theta} \end{aligned} $$

(Here \(\theta\) is the angle made by ladder and the wall.)

\(\theta=\cos ^{-1}\left(\frac{\text { vertical height }}{\text { length of ladder }}\right)\)

$$ \begin{array}{l} =\cos ^{-1}\left(\frac{2.1}{2.5}\right) \\ =32.86^{0} \end{array} $$

So, we have

$$ \begin{aligned} F_{N-\text { wall }} &=\frac{9.8 \times \sin \left(32.86^{\circ}\right)\left(80 \times 2.5+10 \times \frac{2.5}{2}\right)}{2.5 \times \cos \left(32.86^{\circ}\right)} \\ &=538.06 \mathrm{~N} \end{aligned} $$

The frictional force is equal to this force as the frictional force cancels this force. So we

have

$$ F_{f i i}=538.06 \mathrm{~N} $$