# Magnitude of force to keep stick in equilibrium [closed]

Problem statement

A straight and homogenous stick with mass m is pressed against a wall with the force F. The stick is horizontal perpendicular against the wall. Given that the friction between the wall and the stick is μ, determine the horizontal component of F in order for the stick to not fall down.

My thoughts

Forces involved:

We have: The gravitational force mg in the negative y-direction. The normal force from the wall, N(negative x-dreiction). The friction force in the positive y-direction which is f=μN and the force F which acts in the positive x-direction. Thus:

$$\sum F_{x}: F-N=0 \Rightarrow F=N \\ \sum F_{y}: mg-f=mg-\mu N=0 \Rightarrow N=F=\frac{mg}{\mu }$$

$$\boxed{F=\frac{mg}{\mu}}$$

*Correct answer is $\boxed{F=\frac{mg}{2\mu}}$ *

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## closed as off-topic by Brandon Enright, user1504, Waffle's Crazy Peanut, twistor59, Manishearth♦Jun 26 '13 at 22:50

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Does $F_x$ morph into $F$? – DJohnM Jun 16 '13 at 2:59

## 2 Answers

Your mistake is in forgetting that the verictial component of force F holds up half the stick, so the friction force only needs to hold up the other half of the stick's weight, not the whole weight.

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Let the length of the stick be L. Consider moments about the left-most end of the stick. With this choice of axis, neither the applied force F nor the normal reaction force from the wall on the stick contribute to the moments.

Gravity exerts a force of $m \times g$ downward at the centre of the stick, a distance $\frac{L}{2}$ from the left end. This results in a clockwise moment on the stick,$M_{CW}$; $$M_{CW}=m \times g \times \frac{L}{2}$$ Friction exerts a force $F_R$ upward on the right end of the stick, resulting in a counter-clock-wise moment, $M_{CCW}$;$$M_{CCW}=F_R \times L$$For no rotation, the two moments must cancel:$$F_R \times L=m \times g \times \frac{L}{2}$$So, after cancelling L$$F_R=\frac{m \times g}{2}=F_x \times \mu$$ $$F_x=\frac{m \times g}{2 \mu}$$

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