Math Problem Statement

  1. A mail sack with a mass of 85 kg hangs from a vertical rope 2.5 m long. the worker moves the mail sack to the side 0.9 m (measured horizontally) while keeping the length of the rope time is tight. (The attachment point of the upper part of the rope remains in place.) a. What horizontal force is needed to hold the mail sack in the new location? b. When the sack is moved to this location, how much work does the worker do?

Solution

To solve these problems, we need to use concepts from physics, specifically equilibrium and work done by a force.


(a) Horizontal Force Calculation

The situation involves a static equilibrium where the mail sack is held at rest at an angle. The forces acting are:

  • The tension in the rope (T),
  • The weight of the sack (W = mg),
  • The horizontal force (F) exerted by the worker.

The rope forms a triangle with the horizontal displacement x=0.9mx = 0.9 \, \text{m} and the rope length L=2.5mL = 2.5 \, \text{m}. The vertical displacement yy can be found using the Pythagorean theorem:

y=L2x2=(2.5)2(0.9)2=6.250.81=5.44=2.33m.y = \sqrt{L^2 - x^2} = \sqrt{(2.5)^2 - (0.9)^2} = \sqrt{6.25 - 0.81} = \sqrt{5.44} = 2.33 \, \text{m}.

The angle θ\theta between the rope and the vertical can now be found using:

sinθ=xL,cosθ=yL.\sin\theta = \frac{x}{L}, \quad \cos\theta = \frac{y}{L}.

sinθ=0.92.5=0.36,cosθ=2.332.5=0.932.\sin\theta = \frac{0.9}{2.5} = 0.36, \quad \cos\theta = \frac{2.33}{2.5} = 0.932.

At equilibrium, the horizontal force balances the horizontal component of tension:

F=Tsinθ.F = T \sin\theta.

From the vertical forces, tension balances the weight:

Tcosθ=WorT=Wcosθ.T \cos\theta = W \quad \text{or} \quad T = \frac{W}{\cos\theta}.

Substituting W=mg=859.8=833NW = mg = 85 \cdot 9.8 = 833 \, \text{N} and cosθ=0.932\cos\theta = 0.932:

T=8330.932=893.3N.T = \frac{833}{0.932} = 893.3 \, \text{N}.

Now substitute TT and sinθ=0.36\sin\theta = 0.36 into the horizontal force equation:

F=Tsinθ=893.30.36=321.6N.F = T \sin\theta = 893.3 \cdot 0.36 = 321.6 \, \text{N}.


(b) Work Done

The work WW done is the horizontal force multiplied by the horizontal displacement:

W=Fx.W = F \cdot x.

Using F=321.6NF = 321.6 \, \text{N} and x=0.9mx = 0.9 \, \text{m}:

W=321.60.9=289.44J.W = 321.6 \cdot 0.9 = 289.44 \, \text{J}.


Final Answers

  1. Horizontal Force: F=321.6NF = 321.6 \, \text{N}
  2. Work Done: W=289.44JW = 289.44 \, \text{J}

Would you like more detailed steps or have any specific questions?
Here are some related questions to explore further:

  1. How does the angle affect the magnitude of the horizontal force?
  2. What would happen if the rope length were shorter?
  3. How is work affected if the horizontal displacement increases?
  4. What is the potential energy change in the sack during this process?
  5. How can we calculate tension directly using force components?

Tip: When solving static equilibrium problems, always break forces into components and carefully analyze the geometry.

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Math Problem Analysis

Mathematical Concepts

Static Equilibrium
Work and Energy
Trigonometry

Formulas

sin(θ) = opposite/hypotenuse
cos(θ) = adjacent/hypotenuse
F = T * sin(θ)
T = W / cos(θ)
W = F * x

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 11-12