A uniform ladder rests against a smooth wall. The angle between the ladder and the floor is gradually reduced to find the limit at which the ladder can remain at rest.
Adjust conditions:
The sliders adjust the values
床の静止摩擦係数は0.25。静止できる最小の角度は？
壁の静止摩擦係数も0.25なら、静止できる最小の角度はどう変わる？
The equations explain the phenomenon
1An equation of motion is written for each particle.midvidt=Fi+Σ fij……
2The equations are summed over all particles.Σ midvidt=Σ Fi……
3The motion is expressed in terms of the center of mass.M dVGdt=Σ Fi……
4A body can start rotating even when the sum of the forces is 0.F₁+F₂=0……
5Torque is defined.τ=r F sinφ……
6A torque equation is derived from the equations of motion for individual particles.Σ ri × midvidt=Σ ri × Fi……
7The 2 conditions for remaining at rest are derived.Σ F=0，Σ τ=0……
8The conditions are applied to a seesaw.m₁ g ℓ₁=m₂ g ℓ₂……
9Gravity is represented by a single force at the center of mass.Σ mi g xi=M g xG……
10Write the force balance for the ladder.N=M g，f=N₂……
11The reference point is chosen so that 2 of the unknown forces produce 0 torque.N₂ L sinθ=M g (L2) cosθ……
12The limit on static friction is applied.f≦μ N……
13The condition for no slipping is obtained.tanθ≧12μ……
14The case with friction at the wall is examined.tanθ≧1 − μ′μ2μ……
What this chapter has shown
If the sum of the forces is 0, the velocity of the center of mass remains constant. Even an initially stationary rigid body starts rotating if the sum of the torques is not 0. Remaining at rest requires 2 conditions: the sum of the forces is 0, and the sum of the torques is 0.
63.4°Limiting angle for remaining at rest
With a smooth wall and a coefficient of static friction of 0.25 at the floor, the ladder slips if its angle with the floor is reduced below approximately 63.4°. This does not depend on its mass.
In everyday lifeA commonly recommended ladder setup places the foot about 1 unit away from the wall for every 4 units of height. A more slippery floor requires a steeper ladder angle.
An extended object can be analyzed by adding the equations of motion for its constituent particles. The internal forces cancel by Newton's third law, so the motion of the center of mass is determined solely by the net external force. Rotation is governed by torque, and the internal torques also cancel. Equilibrium therefore requires 2 conditions: the sum of the forces is 0, and the sum of the torques is 0. For a ladder, torque balance determines the required frictional force. The ladder begins to slip at an angle where this force exceeds the maximum static frictional force.