Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 4, Problem 1Q
To determine

The force according to Newton, which is needed to produce the horizontal component of an arrow’s velocity.

Expert Solution & Answer
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Answer to Problem 1Q

The horizontal component of an arrow’s velocity is derived from the force applied to the arrow by the bowstring.

Explanation of Solution

Introduction:

According to Newton’s first law, every object continues to be at rest or in the state of uniform motion in a straight line unless an external unbalanced force acts on it.

When an arrow is launched using a bow, the bowstring is pulled, so that energy is stored in it in the form of elastic potential energy. When the string is released, the string exerts a force on the arrow and the arrow gains velocity.

The arrow is initially at rest. The bowstring is pulled and released, keeping the arrow at an angle to the horizontal. If the arrow is launched at an angle, the force exerted on the arrow can be resolved into two components- along the horizontal and the vertical directions. The horizontal component of the force provides the horizontal component of its initial velocity and the vertical component provides the vertical component of its initial velocity. The arrow, thus travels in 2 dimensions, moving both vertically and horizontally.

Conclusion:

Newton’s first law provides an elegant way to explain the two-dimensional motion of the arrow, where the horizontal component of its velocity is provided by the horizontal component of the force exerted on the arrow.

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In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view of the figure. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 202 N, and Charles pulls with force of magnitude 148 N. Note that the direction of Charles' force is not given. What is the magnitude of Betty's force?
Your answer is partially correct. In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force F A of magnitude 205 N, and Charles pulls with force F c of magnitude 184 N. Note that the direction of F c is not given. What is the magnitude of Betty's force F B if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium? Alex Charles 144° Betty (a) Number 304.72 Units N (b) Number 49.1 Units
In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture.
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