Question

A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle of 53 with it.The other end rests on a rough horizontal floor,Find the normal force and the frictional force that the floor exerts on the ladder.

Answer 1

A uniform ladder of mass $10\mathrm{kg}$ leans against a smooth vertical wall making an angle of 53 with it. The other end rests on a rough horizontal floor, Find the normal force and the frictional force that the floor exerts on the ladder.

Solution:

$\mathrm{F_{fr}}$ - frictional force that the floor exerts on the ladder;

$\alpha = 53{}^{\circ}$ - angle which ladder makes with the wall

$\mathrm{N}_{1}$ - reaction force from the floor

$\mathrm{N}_{2}$ - reaction force from the wall

$\mathrm{m} = 10\mathrm{kg}$ - mass of the ladder

L - length of the ladder

Newton's second law for the ladder (the first law of equilibrium):

$$\overrightarrow {\mathrm {F _ {f r}}} + \overrightarrow {\mathrm {m g}} + \overrightarrow {\mathrm {N _ {1}}} + \overrightarrow {\mathrm {N _ {2}}} = \vec {0}$$

Projection of the law on the X-axis:

$$x: N _ {2} - F _ {f r} = 0$$

Projection of the law on the Y-axis:

$$y: N _ {1} - m g = 0$$

Normal force that the floor exerts on the ladder:

$$\mathrm {N} _ {1} = \mathrm {m g} = 1 0 \mathrm {k g} \cdot 9. 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} = 9 8 \mathrm {N}$$

Momentum equation for point A (the second law of equilibrium):

A: $\mathrm{M}_{\mathrm{mg}} + \mathrm{M}_{\mathrm{fr}} + \mathrm{M}_{\mathrm{N}_2} = 0$ (3)

$(\mathrm{M}_{\mathrm{N}_1} = \mathrm{M}_{\mathrm{fr}} = 0,$ because moment arm of this forces is zero)

$\mathrm{M}_{\mathrm{mg}} = -\mathrm{mg}\cdot \frac{\mathrm{L}}{2}\sin \alpha$ (minus sign because of the direction of force)

$$\begin{array}{l} \mathrm{M}_{\mathrm{N}_2} = \mathrm{N}_2 \cdot \mathrm{L} \cos \alpha \\ \rightarrow (3): \end{array}$$ $$\begin{array}{l} \mathrm{N}_2 \cdot \mathrm{L} \cos \alpha - \mathrm{mg} \cdot \frac{\mathrm{L}}{2} \sin \alpha = 0 \\ \mathrm{N}_2 = \frac{\mathrm{mg}}{2} \cdot \frac{\sin \alpha}{\cos \alpha} = \frac{\mathrm{mg} \tan \alpha}{2} \\ \rightarrow (1) \end{array}$$ $$\mathrm{F}_{\mathrm{fr}} = \mathrm{N}_2 = \frac{\mathrm{mg} \tan \alpha}{2} = \frac{10 \, \mathrm{kg} \cdot 9.8 \, \frac{\mathrm{N}}{\mathrm{kg}} \cdot \tan 53{}^\circ}{2} = 65 \, \mathrm{N}$$

Answer: Normal force that the floor exerts on the ladder: $\mathrm{N}_1 = 98 \, \mathrm{N}$; frictional force that the floor exerts on the ladder $\mathrm{F}_{\mathrm{fr}} = 65 \, \mathrm{N}$