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Can you break down the buttom where it comes up with 63 degrees? How can I enter this into a calculator to get the correct answer?

Can you break down the buttom where it comes up with 63 degrees? How can I enter this into a calculator to get the correct answer?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Can you break down the buttom where it comes up with 63 degrees?

How can I enter this into a calculator to get the correct answer?

To answer this question, we'll need to first identify all the forces acting on the hanging mass. To do this, we have to consider the forces acting along the y-axis and x-axis. First, let's consider the forces in the y-direction. Since the mass is not moving in the y-direction, we know that the net force acting on the mass in the vertical direction is zero. We know that the weight of the mass acts downward, and the vertical component of the string's tension acts upward. ΣFy = Tsin(0) — mg = 0 Tsin(0) = mg Next, let's look at the forces acting on the mass in the horizontal direction. Since the mass is moving in a uniform circular motion, we know that there is a centripetal force associated with this. Moreover, we know that the x-direction of the tension in the string is acting in the x- direction. Thus, this component of the string's tension must be providing the centripetal force. ΣF = Tcos (0) tan(Ⓒ) Now, if we divide these two equations, we can cancel out the terms for tension and mass. Tsin(e) Tcos (0) = e rg v2 = tan¯¹(. Ꮎ ~ 63° mv² mg mv² T Further rearrangement of the above expression allows us to solve for the angle. rg = tan -¹ (2) T (5.0 m) (9.8m) (5 m)2
Transcribed Image Text:To answer this question, we'll need to first identify all the forces acting on the hanging mass. To do this, we have to consider the forces acting along the y-axis and x-axis. First, let's consider the forces in the y-direction. Since the mass is not moving in the y-direction, we know that the net force acting on the mass in the vertical direction is zero. We know that the weight of the mass acts downward, and the vertical component of the string's tension acts upward. ΣFy = Tsin(0) — mg = 0 Tsin(0) = mg Next, let's look at the forces acting on the mass in the horizontal direction. Since the mass is moving in a uniform circular motion, we know that there is a centripetal force associated with this. Moreover, we know that the x-direction of the tension in the string is acting in the x- direction. Thus, this component of the string's tension must be providing the centripetal force. ΣF = Tcos (0) tan(Ⓒ) Now, if we divide these two equations, we can cancel out the terms for tension and mass. Tsin(e) Tcos (0) = e rg v2 = tan¯¹(. Ꮎ ~ 63° mv² mg mv² T Further rearrangement of the above expression allows us to solve for the angle. rg = tan -¹ (2) T (5.0 m) (9.8m) (5 m)2
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