A ladder, leaning against a wall, makes a 60° angle with the ground. When it is more likely to slip: when a person stands on the ladder near the top or near the bottom? Explain.

Force may be defined as the product of the object’s acceleration and mass. If the value of the net force working on an object is equal to zero, then the object is assumed to be in translational equilibrium.

The free-body diagram for the given situation can be drawn as:

Here,\(Mg\)is the force of the person,\(mg\)is the force of the ladder,\({F_{{\rm{Gy}}}}\)is the vertical component of the frictional force at the ground,\({F_{{\rm{Gx}}}}\)is the horizontal component of the frictional force at the ground,\({F_{\rm{W}}}\)is the reaction force exerted by the wall on the ladder,\({d_{\rm{x}}}\)is the horizontal distance of the person, and\({d_{\rm{y}}}\)is the vertical distance of the person.

The person's force \(\left( {Mg} \right)\) acting downward on the ladder gives a torque in the clockwise direction about the junction point with the ground, with lever arm \({d_{\rm{x}}}\). The only force creating torque in the counterclockwise direction about that same point is the \({F_{\rm{w}}}\).

Since the ladder is in a state of equilibrium, \({F_{\rm{w}}}\) will be the identical magnitude as \({F_{{\rm{Gx}}}}\). Since \({F_{{\rm{Gx}}}}\) has the highest value, \({F_{\rm{w}}}\) will have the corresponding highest value, and \({F_{\rm{w}}}\) will have the maximum counterclockwise torque that it can exert.

As the person ascends the ladder, his lever arm becomes longer; so the torque due to his weight becomes higher.

Finally, if the torque produced by the person is larger than the maximum torque produced by\({F_{\rm{w}}}\), the ladder will begin to slip, and it will not be in the state of equilibrium.