A person is trying to lift a ladder of mass \(15 \mathrm{kg}\) and length $8.0 \mathrm{m} .$ The person is exerting a vertical force on the ladder at a point of contact \(2.0 \mathrm{m}\) from the center of gravity. The opposite end of the ladder rests on the floor. (a) When the ladder makes an angle of \(60.0^{\circ}\) with the floor, what is this vertical force? (b) A person tries to help by lifting the ladder at the point of contact with the floor. Does this help the person trying to lift the ladder? Explain.

Draw a diagram of the ladder with all the forces acting on it. Label important lengths (e.g. length of the ladder, distance of the center of gravity from the floor, and distance of the lifting force from the floor) and angles.

We can break down the forces acting on the ladder into their horizontal and vertical components. We have: - The weight of the ladder acting downwards at the center of gravity: \(mg\) - The vertical force exerted by the person lifting the ladder: \(F_v\) - The normal force from the floor acting on the end of the ladder: \(N\)

Torque is calculated as the force multiplied by the perpendicular distance to the pivot point. The pivot point, in this case, is the end of the ladder resting on the floor. We'll calculate the torque due to \(F_v\) and the torque due to the weight of the ladder in vertical direction. Torque due to \(F_v\): Since the force is vertical and the distance from the pivot point to the point of contact (where the force is exerted) is 2m, we have: \(τ_{F_v} = F_v \times (2\,\mathrm{m})\) Torque due to the weight of the ladder: Let's call \(L\) the distance from the pivot point to the center of gravity, which is half the length of the ladder (\(L = 8.0\,\mathrm{m} / 2 = 4.0\,\mathrm{m}\)). Since the weight of the ladder acts downwards, the torque is calculated using the perpendicular distance to the pivot point, which can be found using trigonometry: \(τ_{mg} = mg \times (L \sin(60^{\circ}))\)

Since the ladder is in equilibrium, the torques balance each other: \(τ_{F_v} = τ_{mg}\) Substituting the expressions for the torques: \(F_v \times (2\,\mathrm{m}) = mg \times (4.0\,\mathrm{m} \sin(60^{\circ}))\) Solve for \(F_v\): \(F_v = \frac{mg (4.0\,\mathrm{m} \sin(60^{\circ}))}{2\,\mathrm{m}}\) Substitute the values of mass \((15\,\mathrm{kg})\) and gravitational acceleration \((9.81\,\mathrm{m/s^2})\): \(F_v = \frac{(15\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4.0\,\mathrm{m} \sin(60^{\circ}))}{2\,\mathrm{m}} \approx 204.11\,\mathrm{N}\) (a) The vertical force exerted by the person is approximately \(204.11\,\mathrm{N}\).

If a second person tries to help by lifting the ladder at the point of contact with the floor, the additional force they provide would either introduce a vertical or horizontal component or both. Since the normal force from the floor keeps the ladder from sinking into the ground, a vertical force from the second person at this point would have no effect on the overall vertical force needed by the first person. However, if the second person applies a horizontal force, it could potentially change the angle between the ladder and the floor, which could in turn change the torque balance, either increasing or decreasing the required force exerted by the first person. (b) Lifting the ladder at the point of contact with the floor in vertical direction does not help the first person. The second person should apply a horizontal force in order to help the first person efficiently.