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CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

Short Answer

Expert verified
(a) 67.2 N, (b)(i) 0 J, (ii) 134.4 J

Step by step solution

01

Identify Known Quantities

We know that the mass of the mail bag is 120 kg, the length of the rope is 3.5 m, and the horizontal displacement is 2.0 m. We'll use these values to determine other necessary physical parameters.
02

Draw the Position Diagram

Visualize the situation. The bag originally hangs vertically. When it's displaced sideways by 2.0 m while keeping the rope taut, the bag forms a right triangle with the rope being the hypotenuse, the vertical displacement as one side, and horizontal displacement as the base.
03

Calculate the Vertical Displacement

Using Pythagorean theorem in the right triangle, solve for the vertical displacement (height the bag is lifted, h): \\( h = \sqrt{3.5^2 - 2.0^2} \).
04

Consider Forces acting on the Bag

The bag is in equilibrium, hence the horizontal force exerted by the worker should balance the horizontal component of tension in the rope. Gravity acts downward and tension acts along the rope. Use trigonometry to relate these forces.
05

Calculate Horizontal Force

Using trigonometric ratios: tan(\( \theta \)) = \( \frac{{2.0}}{h} \), calculate \( \theta \) and then the horizontal component of tension T using T = mg / cos(\( \theta \)). Set horizontal force F = T * sin(\( \theta \)).
06

Determine Work Done by the Rope

Since the rope is assumed to be massless and does not contribute work because it only changes the direction of tension without pulling or lifting.
07

Calculate Work Done by the Worker

The worker does work in moving the bag horizontally. Calculate it as: \( \text{Work} = F \cdot d \cdot \cos(0) = F \cdot 2.0 \). Use the horizontal force previously found.

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Key Concepts

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

Pythagorean theorem
The Pythagorean theorem is a fundamental principle in mathematics and physics, essential for problems involving right triangles. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, it is expressed as:\[ c^2 = a^2 + b^2 \]Here, \(c\) represents the hypotenuse, while \(a\) and \(b\) represent the other two sides. In our problem, we apply this theorem to determine the vertical displacement of the mail bag when it is displaced sideways by 2.0 m. The rope forms the hypotenuse (3.5 m), the horizontal displacement is one leg (2.0 m), and the vertical displacement is the unknown side. By rewriting the equation to solve for this missing height (h), we get:\[ h = \sqrt{3.5^2 - 2.0^2} \]This calculation allows us to find how much the mail bag is lifted vertically, which helps us in understanding the mechanics of the entire setup.
trigonometry in physics
Trigonometry plays a crucial role in solving physics problems, especially those involving forces and angles. Our problem requires an understanding of how trigonometric functions like sine, cosine, and tangent relate to the components of forces in a right triangle setup. When the mail bag is displaced, it forms a triangle where:- The rope acts as the hypotenuse.- The vertical and horizontal displacements are the other sides.To find the angle \(\theta\) that the rope makes with the vertical, we use the tangent function:\[ \tan(\theta) = \frac{2.0}{h} \]Calculating \(\theta\) lets us break down the tension in the rope into its vertical and horizontal components. This is essential because:- Vertical component is balanced by the gravitational force \(mg\).- Horizontal component is what the postal worker needs to apply to keep the bag still.These trigonometric calculations simplify how we can analyze and solve the problem using forces and angles.
horizontal force calculation
The horizontal force calculation is crucial to ascertain how much force the worker needs to exert to hold the mail bag in its displaced position. Once we have the angle \(\theta\), we resolve the tension in the rope into its components. The tension \(T\) in the rope is related to gravity as it's the force countering the weight of the bag:\[ T = \frac{mg}{\cos(\theta)}\]Where:- \(m\) is the mass of the mail bag (120 kg).- \(g\) is the acceleration due to gravity, approximately 9.8 \(\text{m/s}^2\).Next, since the mail bag is at rest horizontally, the horizontal component of the tension must balance the horizontal force the worker applies. Using the sine function, we have:\[ F = T \cdot \sin(\theta)\]Thus, this horizontal force \(F\) needed to maintain the position is crucial for understanding equilibrium in this statics problem. This approach highlights the blend of physics and trigonometry to solve real-world problems efficiently.

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Most popular questions from this chapter

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EP A small rock with mass 0.20 \(\mathrm{kg}\) is released from rest at point \(A,\) which is at the top edge of a large, hemispherical bowl with radius \(R=0.50 \mathrm{m}\) (Fig. E7. 9\()\) . Assume that the size of the rock is small compared to \(R,\) so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point \(A\) to point \(B\) at the bottom of the bowl has magnitude 0.22 J. (a) Between points \(A\) and \(B\) , how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point \(B ?(\mathrm{c})\) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point \(B,\) what is the normal force on it due to the bottom of the bowl?

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