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Chapter 4: Problem 16

The force exerted by the wind on the sails of a sailboat is \(390 \mathrm{~N}\) north. The water exerts a force of \(180 \mathrm{~N}\) east. If the boat (including its crew) has a mass of \(270 \mathrm{~kg}\), what are the magnitude and direction of its acceleration?

Short Answer

Expert verified
The magnitude of the acceleration is approximately \(2.19 \, \mathrm{ms}^{-2}\) and the direction is \(24.6^\circ\) east of north.

Step by step solution

01

Determine the Resultant Force

The resultant force acting on the ship can be calculated by considering each force as a vector. The force due to the wind is pointing towards North (\(F_{wind} = 390 N\)), whereas the force due to the water is pointing towards the East (\(F_{water} = 180 N\)). These forces act at right angles to each other thus the magnitude of the resultant force (\(F_{R}\)) can be found by using Pythagoras' theorem: \( F_{R} = \sqrt{F_{wind}^2 + F_{water}^2} = \sqrt{390^2 + 180^2} = \sqrt{210,600}\).
02

Total Acceleration

The acceleration (\(a\)) of the boat can be found by using Newton's second law, which is stated as \( F = ma \). Rearranging for \( a \), we have \( a = \frac{F}{m} \). We plug in the values of \( F_{R} \) and the given mass into this equation to find the acceleration. \( a = \frac{\sqrt{210,600}}{270} \).
03

Direction of Acceleration

The direction of the acceleration can be found by determining the angle east of north. This can be calculated by finding the angle \( \theta \) using the equation \( \theta = \tan^{-1}(\frac{F_{water}}{F_{wind}}) \). Substituting the known values gives \( \theta = \tan^{-1}(\frac{180}{390}) \).

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Key Concepts

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

Resultant Force Calculation
Understanding the concept of resultant force is crucial when analyzing the motion of objects influenced by multiple forces. A resultant force is effectively the single force that represents the vector sum of all the forces acting on an object, defining the net force that dictates the object's acceleration. In the provided exercise, the sailboat experiences two perpendicular forces: a wind force of 390 N north and water resistance of 180 N east.

To find the magnitude of the resultant force, we use Pythagoras' theorem, which is applicable here since the forces are at a right angle to each other. The equation for the resultant force (\(F_R\)) becomes: \[ F_{R} = \( \( \( \(F{\_wind}^{2} + F{\_water}^{2} \( \)\)\)\) \) = \( \( \( \( \sqrt{390^{2} + 180^{2}} \( \)\)\)\) \) = \( \( \( \( \sqrt{210,600} \( \)\)\)\) \)\]Making sense of this calculation is key to predicting the boat's movement.
Newton's Second Law of Motion
At the heart of motion analysis is Newton's second law of motion, which provides a relationship between the forces acting on an object, its mass, and the resulting acceleration. It is succinctly expressed in the equation \( F = ma \) where \( F \) represents the net force acting on the object, \( m \) is the mass of the object, and \( a \) is the acceleration.

Applying this law to the exercise, once we have the magnitude of the resultant force, we determine the acceleration of the boat by rearranging the equation to solve for \( a \) 鈥� \( a = \frac{F}{m} \). Plugging in the values from the resultant force and the boat's mass, we can find out how fast the boat will accelerate in response to the combined forces of wind and water.
Vector Addition in Physics
Vectors are mathematical representations of quantities that have both a magnitude and a direction. In physics, vector addition is commonly used to combine multiple forces acting on an object to find a single resultant force. Since forces are vector quantities, they follow the rules of vector addition.

For instance, if two forces are perpendicular, as in the case of the sailboat, their vector addition is akin to the sides and hypotenuse of a right triangle. Therefore, to obtain the direction of the resultant force, we use trigonometry鈥攕pecifically the tangent function鈥攚hich relates the sides of a right triangle to its angles. The angle \( \theta \) east of north that gives us the direction of the total acceleration can be found using the equation \( \theta = \tan^{-1}(\frac{F_{water}}{F_{wind}}) \). The calculated angle helps us understand not only how fast the boat is accelerating but in what direction it will travel.

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Most popular questions from this chapter

A crate of mass \(45.0 \mathrm{~kg}\) is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is \(0.350\), and the coefficient of kinetic friction is \(0.320\). (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force \(F\) exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of \(1.50 \mathrm{~m}\) and a mass of \(61.0 \mathrm{~kg}\). If this fish is moving upward at \(3.00 \mathrm{~m} / \mathrm{s}\) as its head first breaks the surface and has an upward speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after two-thirds of its length has left the surface, assume constant acceleration and determine (a) the salmon's acceleration and (b) the magnitude of the force \(F\) during this interval.

A \(15.0\)-lb block rests on a horizontal floor. (a) What force does the floor exert on the block? (b) A rope is tied to the block and is run vertically over a pulley. The other end is attached to a free-hanging \(10.0\)-lb object. What now is the force exerted by the floor on the 15.0-lb block? (c) If the 10.0-lb object in part (b) is replaced with a \(20.0\)-lb object, what is the force exerted by the floor on the \(15.0\)-lb block?

A frictionless plane is \(10.0 \mathrm{~m}\) long and inclined at \(35.0^{\circ} .\) A sled starts at the bottom with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed \(v_{i}\). Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled.

A boat moves through the water with two forces acting on it. One is a \(2000-\mathrm{N}\) forward push by the water on the propeller, and the other is a \(1800-\mathrm{N}\) resistive force due to the water around the bow. (a) What is the acceleration of the \(1000-\mathrm{kg}\) boat? (b) If it starts from rest, how far will the boat move in \(10.0 \mathrm{~s}\) ? (c) What will its velocity be at the end of that time?
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