Sliding Ladder

In static equilibrium, the sum of the external forces acting on the object must be zero. This also implies that if there is any external torque acting on the object, it must also sum to zero. In this demonstration, a ladder with a hanging mass is resting against a wall in static equilibrium at some angle. At the base of the ladder, there is a rotating block with surfaces of variable coefficients of friction. The forces acting on the base are the force due to gravity pointing downward, the normal force pointing upward, and the force due to friction pointing towards the wall. The force due to friction is equal to the product of the coefficient of friction and normal force, which are independent of the ladder’s angle with respect to the ground. At the top of the ladder, there is a normal force pointing away from the wall, which is balanced by the force due to friction at the bottom of the ladder. In static equilibrium, the net torque must also sum to zero, and the axis of rotation can be considered at the base of the ladder. The torque directed in the clockwise direction due to the normal force at the top of the ladder is counterbalanced by a torque directed in a counterclockwise direction due to the weight of the ladder and hanging mass. When the mass is moved up the rungs of the ladder, this increases its radius from the axis of rotation, and thus the torque due to the hanging mass increases. This increases the normal force acting at the top of the ladder; however, the counteracting force of friction acting at the base of the ladder remains fixed. If the ladder is tilted at a large enough angle with respect to the ground and the hanging mass is past a critical distance from the base of the ladder, the ladder will fall to the ground.