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Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 5, Problem 90

A train traveling at a constant speed rounds a curve of radius 215 m. A lamp suspended from the ceiling swings out to an angle of 18.5° throughout the curve. What is the speed of the train? [Hint: See Example 4–15.]

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Identify the forces acting on the lamp. The lamp is subject to two forces: the tension in the string (T) and the gravitational force (mg). The tension can be broken into two components: a vertical component (Tcosθ) that balances the gravitational force and a horizontal component (Tsinθ) that provides the centripetal force.
Write the equations for the forces. Vertically, the forces are balanced: Tcosθ = mg. Horizontally, the centripetal force is provided by the horizontal component of tension: Tsinθ = m(v²/r), where v is the speed of the train and r is the radius of the curve.
Eliminate T from the equations. Divide the horizontal force equation by the vertical force equation to eliminate T: (Tsinθ) / (Tcosθ) = (m(v²/r)) / (mg). This simplifies to tanθ = v² / (rg).
Solve for the speed of the train (v). Rearrange the equation tanθ = v² / (rg) to isolate v: v = √(rg * tanθ).
Substitute the known values into the equation. Use r = 215 m, g = 9.8 m/s², and θ = 18.5° (convert to radians if necessary). This will allow you to calculate the speed of the train.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of the circle. It is essential for maintaining circular motion and is calculated using the formula F_c = m*v^2/r, where m is mass, v is velocity, and r is the radius of the circle. In this scenario, the train's speed can be determined by analyzing the forces acting on the lamp as it swings outward.
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Angle of Deviation

The angle of deviation refers to the angle at which an object, such as the lamp, swings away from its vertical position due to the forces acting on it. In this case, the lamp swings out to an angle of 18.5° as the train rounds the curve, indicating a balance between gravitational force and the horizontal component of the tension in the lamp's support. This angle is crucial for calculating the effective forces and ultimately the speed of the train.
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Trigonometric Relationships

Trigonometric relationships, particularly involving sine and cosine, are used to resolve forces acting at angles. In this problem, the angle of 18.5° can be used to relate the vertical and horizontal components of the forces acting on the lamp. By applying these relationships, one can derive the speed of the train from the geometry of the situation and the forces involved.
Related Practice
Textbook Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude FT = 240N to the skier (θ = 0°)?

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Textbook Question

The 70.0-kg climber in Fig. 5–53 is supported in the 'chimney' by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that the static friction forces are both at their maximum. Ignore his grip on the rope.

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Textbook Question

A coffee cup on the horizontal dashboard of a car slides forward when the driver decelerates from 45 km/h to rest in 3.5 s or less, but not if she decelerates in a longer time. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are level (horizontal).

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Textbook Question

A car starts rolling down a 1-in-4 hill (1-in-4 means that for each 4 m traveled along the road, the elevation change is 1 m). How fast is it going when it reaches the bottom after traveling 55 m? Assume an effective coefficient of rolling friction equal to 0.10.

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Textbook Question

The position of a particle moving in the xy plane is given by r\(\overrightarrow{r}\) = (2.0m) cos [(3.0 rad/s)t ] i^\(\hat{i}\) +(2.0m) sin [(3.0 rad/s)t ] j^\(\hat{j}\), where r is in meters and t is in seconds. Calculate the velocity and acceleration vectors as functions of time.

Textbook Question

A 28.0-kg block is connected to an empty 2.00-kg bucket by a cord running over a frictionless pulley (Fig. 5–56). The coefficient of static friction between the table and the block is 0.42 and the coefficient of kinetic friction between the table and the block is 0.34. Sand is gradually added to the bucket until the system just begins to move. Calculate the acceleration of the system.