Question

A 11.0-kg monkey climbs a uniform ladder with weight w = 1.10 ✕ 102 N and length L = 3.70 m as shown in the figure below. The ladder rests against the wall at an angle of theta = 60°. The upper and lower ends of the ladder rest on frictionless surfaces, with the lower end fastened to the wall by a horizontal rope that is frayed and that can support a maximum tension of only 80.0 N. A monkey climbs a ladder of length L that is leaning to the right against a wall. A horizontal rope connects the base of the ladder to the base of the wall. The ladder makes an acute angle theta counterclockwise from the horizontal. (a) Draw a force diagram for the ladder. (Submit a file with a maximum size of 1 MB.) (b) Find the normal force exerted on the bottom of the ladder. N (c) Find the tension in the rope when the monkey is two-thirds of the way up the ladder. N (d) Find the maximum distance d that the monkey can climb up the ladder before the rope breaks. m (e) If the horizontal surface were rough and the rope were removed, how would your analysis of the problem be changed and what other information would you need to answer Parts (c) and (d)?

Answer 1

(a) The force diagram for the ladder includes the weight acting downward at the center, the normal force exerted by the floor, the tension in the rope, and the frictional force at the bottom. There's also the force exerted by the wall at the upper end, perpendicular to the wall, and the horizontal component of the tension in the rope.

(b) The normal force exerted on the bottom of the ladder is 684.4 N.

(c) The tension in the rope when the monkey is two-thirds of the way up the ladder is 225.4 N.

(d) The maximum distance the monkey can climb up the ladder before the rope breaks is 2.13 m.

(e) If the horizontal surface were rough and the rope were removed, the analysis would need to consider the frictional force at the bottom. Additional information required would include the coefficient of friction between the ladder and the ground to determine the maximum frictional force.

Step-by-step explanation:

In the force diagram, the normal force exerted on the bottom of the ladder (b) can be determined using the equilibrium condition in the vertical direction, yielding N = mg + Tsin(θ), where m is the monkey's mass, g is the gravitational acceleration, T is the tension in the rope, and θ is the angle of the ladder.

To find the tension in the rope when the monkey is two-thirds of the way up (c), we use the torque equation. Summing torques about the bottom of the ladder gives T = (2/3)mgLsin(θ)/(1/3L).

For the maximum distance the monkey can climb before the rope breaks (d), we set the tension in the rope equal to the maximum tension it can support (80 N) and solve for the distance.

If the surface were rough and the rope removed (e), the frictional force would come into play at the bottom of the ladder. The analysis would require the coefficient of friction (μ) and would involve setting up equations that balance the forces in both the vertical and horizontal directions.

Answer 2

(a) Force Diagram: In the force diagram, the ladder experiences weight downward, countered by the normal force, tension in the rope pulling horizontally, and friction opposing motion.

(b) Normal Force (N): The normal force is approximately 55.0 N, acting perpendicular to the surface.

(c) Tension in the Rope (T): The tension in the rope, when the monkey is two-thirds up, is approximately 95.1 N, pulling horizontally.

(d) Maximum Climbing Distance (d): The monkey can climb up to approximately 0.76 m before the rope breaks, considering the ladder's equilibrium.

(e) Surface Friction and Rope Removal: Analysis of surface friction requires additional information on the coefficient of friction. Without the rope, the ladder wouldn't be horizontally stabilized.

Given Data:

Monkey's weight (w) = 1.10 × 10^2 N

Length of the ladder (L) = 3.70 m

Angle with the horizontal (θ) = 60 degrees

Maximum tension in the rope (T_max) = 80.0 N

(a) Force Diagram:

Weight of the ladder (w) acts vertically downward.

Normal force (N) acts perpendicular to the surface.

Tension in the rope (T) pulls horizontally.

Friction (f) opposes the ladder's motion.

(b) Normal Force (N):

N = w × cos(θ)

N = (1.10 × 10^2 N) × cos(60 degrees)

N ≈ 55.0 N

(c) Tension in the Rope (T) when the monkey is two-thirds up:

T = w × sin(θ)

T = (1.10 × 10^2 N) × sin(60 degrees)

T ≈ 95.1 N

(d) Maximum Climbing Distance (d):

To find d, consider the ladder's equilibrium:

T_max = w × sin(θ)

80.0 N = (1.10 × 10^2 N) × sin(60 degrees)

Sin(60 degrees) ≈ 0.866

d = T_max / (w × sin(θ))

d ≈ 80.0 N / ((1.10 × 10^2 N) × 0.866)

d ≈ 0.76 m

(e) Surface Friction and Rope Removal:

To analyze surface friction, additional information about the coefficient of friction and the roughness of the surface is needed. Without the rope, the ladder wouldn't be stabilized horizontally.