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A ladder is $13\ \mathrm m$ long and weights $300\ \mathrm N$, rests against a smooth wall at height $12\ \mathrm m$ above the floor. The floor is rough. Find the frictional force and the normal force exerted on ladder by the floor and the normal force exerted on the ladder by the wall.

The answer provided in the textbook is $F_\text{wall} = 300\ \mathrm N$ and $F_\text{floor} = 62.5\ \mathrm N$. Only these two answers are provided and I believe they are the normal forces.

Shouldn’t they be swapped as the normal force by the floor should cancel out with the weight and therefore be $300\ \mathrm N$?

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  • $\begingroup$ That seems correct as the only vertical force on the ladder is from the floor. $\endgroup$ – Joe Iddon Mar 11 at 20:16
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Schematic of the static ladder depicting lengths and forces acting on it.

In the schematic figure, the static ladder, wall and floor are depicted by the hypotenuse and sides of a right triangle and the forces on it are depicted in red. Newton's laws imply that the reaction forces $F_\text{wall}$ and $F_\text{friction}$ satisfy $$F_\text{floor} = W$$ $$\text{ and} \; F_\text{wall} = F_\text{friction},$$ so that $F_\text{floor} = 300\ \mathrm N$. Using these facts and calculating torques at the point of contact of the ladder with the wall implies that $$F_\text{floor} \cdot \frac{5}{13} \cdot 13 = W \cdot \frac{5}{13} \cdot 13 = W \cdot \frac{5}{13} \cdot \frac{13}{2} + F_\text{friction} \cdot \frac{12}{13} \cdot 13,$$ so that $F_\text{wall} = 62.5\ \mathrm N$.

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    $\begingroup$ The figure has a type. The horizontal dimension should be 5 m instead of 12 m. I did not check your math, but I do see values like 5/13 and 12/13, so your answer is probably correct. $\endgroup$ – JohnHoltz Mar 17 at 15:12
  • $\begingroup$ Good catch @JohnHoltz. I have edited the figure to correct the error. $\endgroup$ – kb314 Mar 17 at 15:50
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Yes, swap the answers. Friction with the floor is equal and opposite to normal at the wall.

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