How is static equilibrium formed?

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How is static equilibrium formed?

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Answer 1 · 2015-12-05T05:01:36

When the sum of all forces is $0$ $0$ , and the object in question doesn't move at all, it is in static equilibrium.

A classic example is a man standing on a ladder propped up against a wall.

where $m$ $m$ is the mass of the man, $M$ $M$ is the mass of the ladder, $g$ $g$ is the gravitational constant, $F_{s}$ $F_s$ is the static friction force, $F_{N}$ $F_N$ is the normal force, and $θ$ $theta$ is the angle with the horizontal. Ladder length is $L$ $L$ .

So, you have, with CW torque = positive, and downward and rightward force being positive:

$\sum F_{x} = F_{N 1} - F_{s 2} = 0$ $sum F_x = F_(N1) - F_(s2) = 0$

$\sum F_{y} = (M + m) g - F_{N 2} - F_{s 1} = 0$ $sum F_y = (M+m)g - F_(N2) - F_(s1) = 0$

$\sum τ_{x} = τ_{F_{N 1}} + τ_{F_{s 2}} = 0$ $sum tau_x = tau_(F_(N1)) + tau_(F_(s2)) = 0$

$\sum τ_{y} = τ_{F_{s 1}} - τ_{F_{N 2}} - τ_{m g} - τ_{M g} = 0$ $sum tau_y = tau_(F_(s1)) - tau_(F_(N2)) - tau_(mg) - tau_(Mg) = 0$

CHALLENGE: If you only knew $L$ $L$ , $θ$ $theta$ , $μ_{s 1}$ $mu_(s1)$ , $μ_{s 2}$ $mu_(s2)$ , $m$ $m$ , and $M$ $M$ , can you find expressions for each force in terms of the known variables? Assume that the man is halfway up along the ladder and that the ladder's center of mass is halfway along the ladder.

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How is static equilibrium formed?

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