An archer draws a bow, getting ready to shoot a 55.0 gram arrow. The tension in the bowstring is 424 N, and it is bent at the angles 30 degrees
a)What is the magnitude of the "drawing" force holding the string at rest?
b)If the drawing force is removed (by the archer releasing the string to shoot the arrow), what will be the acceleration of the arrow? All forces can be neglected on the arrow except the forces being exerted by the bowstring.
F = 2 * 424 * sin 30 = 424
F = m A
A = 424/.055 = 7709 m/s^2
To find the answers to these questions, we can make use of Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
a) To find the magnitude of the "drawing" force holding the string at rest, we need to determine the horizontal component of the tension in the bowstring. This can be found using trigonometry.
Given that the tension in the bowstring is 424 N and it is bent at an angle of 30 degrees, we can calculate the horizontal component of the tension (drawing force) using the equation:
Drawing force = Tension x cos(angle)
Drawing force = 424 N x cos(30 degrees)
Drawing force = 424 N x 0.86603 ≈ 367.0 N
Therefore, the magnitude of the "drawing" force holding the string at rest is approximately 367.0 N.
b) To find the acceleration of the arrow after the drawing force is removed, we need to consider Newton's second law once again. In this case, the net force acting on the arrow is provided solely by the tension in the bowstring.
The net force (F_net) acting on the arrow can be calculated using the equation:
F_net = Mass x Acceleration
Since the force exerted by the bowstring is now the net force acting on the arrow, we can rewrite the equation as:
Drawing force = Mass x Acceleration
Rearranging the equation, we can solve for the acceleration:
Acceleration = Drawing force / Mass
Acceleration = 367.0 N / 0.055 kg ≈ 6672 m/s²
Therefore, the acceleration of the arrow, neglecting all other forces, would be approximately 6672 m/s².