91Ó°ÊÓ

QIC I S A painter climbs a ladder leaning against a smooth wall. At a certain height, the ladder is on the verge of slipping. (a) Explain why the force exerted bythe vertical wall on the ladder is horizontal. (b) If the ladder of length \(L\) leans at an angle \(\theta\) with the horizontal, what is the lever arm for this horizontal force with the axis of rotation taken at the base of the ladder? (c) If the ladder is uniform, what is the lever arm for the force of gravity acting on the ladder? (d) Let the mass of the painter be \(80 \mathrm{~kg}, L=4.0 \mathrm{~m}\), the ladder's mass be \(30 \mathrm{~kg}, \theta=53^{\circ}\), and the coefficient of friction between ground and ladder be \(0.45\). Find the maximum distance the painter can climb up the ladder.

Step by step solution

01

Understanding Force Orientation

The painter is leaning against a smooth wall. Since the wall is smooth, the normal force (force exerted by the wall on the ladder) will be perpendicular to it, hence it would be horizontal.

02

Lever Arm for Horizontal Force

The lever arm for a force is the perpendicular distance from the axis of rotation to the line along which the force acts. In this case, the ladder makes an angle \(\theta\) with the horizontal, so the lever arm for the vertical force (Lever arm_v) would be \(L \sin (\theta)\).

03

Lever Arm for the Force of Gravity

The ladder is uniform, so its center of mass is at its midpoint. The force of gravity acts downward from this point. Hence, the lever arm for the force of gravity (Lever arm_g) is half the ladder length times sin of the angle, hence \(0.5L \sin (\theta)\).

04

Maximum Distance Calculation

We know that the torque due to the acting forces must balance. Since the ladder is on the point of slipping, the frictional force equals the maximum static frictional force, and the sum of torques equals to 0. Let's denote the max distance the painter can climb as \(h\). Now, considering how torques balance out: \n \\(mg (0.5L \sin (\theta)) + 80g (h \sin (\theta)) = \mu (80g + mg)(L \sin (\theta))\), where \(\mu\) is the coefficient of friction, and \(m \& g\) denote the ladder's mass and gravitational force respectively. Solving for \(h\) we obtain: \n \\(h = \frac{\mu (80 + m)(L) - 0.5mL}{80}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force Exertion

When a painter leans a ladder against a smooth wall, understanding the force exertion is crucial for safety and balance. The wall is smooth, which means there’s no friction between it and the ladder. This lack of friction means the only force the wall can exert on the ladder is horizontal.
This force is perpendicular to the wall because that’s the way "normal forces" work—they act at right angles.

• The normal force here is horizontal and acts to push the ladder away from the wall.
• There’s no vertical component from the wall because friction isn’t at play.

Remember that this horizontal force is crucial in preventing the ladder from slipping, providing balance against gravitational forces pulling the ladder down.

Lever Arm Calculations

Calculating the lever arm helps us understand how forces cause rotational movement or torque. When a force acts at a distance from an axis of rotation, it creates torque, which is what turns or rotates objects.
For the painter’s ladder, the axis of rotation is at the ladder’s base. The lever arm for the horizontal force is the distance from this axis to where the force acts.

• This distance is determined by the angle \( \theta \) of the ladder with the ground.
• The formula for this lever arm is \( L \sin(\theta) \), where \( L \) is the length of the ladder.

Understanding this distance helps us figure out how much torque is generated, which is key in ensuring balance and preventing movement.

Friction and Motion

Friction is the force resisting the relative motion of surfaces in contact. It’s what prevents the ladder from slipping straight to the ground. In this problem, we have to consider the friction between the ladder's base and the ground.
The frictional force is influenced by several factors:

• The coefficient of friction \( \mu \) between the two surfaces, which here is 0.45.
• The normal force exerted on the ladder’s base, which is the combined weight of the painter and the ladder.

The ladder will remain stable as long as the torque from the frictional force can balance the torques from gravity and the painter’s weight. Maximum static friction is vital to calculate how far the painter can ascend without causing slipping.

Torque and Balance

For the ladder to be in balance while the painter climbs, the sum of torques—rotational forces around an axis—must equal zero. This ideal balance of torques ensures the ladder doesn’t rotate or slip.
We calculate the torque from each force acting around the base of the ladder (axis of rotation):

• Gravitational torque from the ladder’s weight: \( mg(0.5L \sin(\theta)) \), with \( m \) being ladder’s mass.
• Painter’s torque as he climbs: \( 80g(h \sin(\theta)) \), with \( h \) as the height he climbs.
• Torque from friction: \( \mu(80g + mg)(L \sin(\theta)) \), providing balance with the other torques.

Too much torque from the painter’s weight will overcome friction, leading to slipping. Solving the equation helps find the max height \( h \) the painter can reach safely.