(d) Assume that the magnitude of the tension in the rope is fixed but that the angle may be varied. For what value of 0 would the resulting horizontal acceleration of the crate be maximized?
(d) Assume that the magnitude of the tension in the rope is fixed but that the angle may be varied. For what value of 0 would the resulting horizontal acceleration of the crate be maximized?
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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id like to see part d plz!
![**1. Topic: Motion of a Crate Being Pulled Across a Horizontal Surface**
This exercise addresses the motion of a crate that is being pulled across a horizontal floor by a rope. In the diagram provided, the following variables are defined:
- \( m \): Mass of the crate
- \( \mu \): Coefficient of kinetic friction between the crate and the floor
- \( F_T \): Tension in the rope
- \( \theta \): Angle of the rope above the horizontal
**Diagram Explanation:**
The diagram shows a crate of mass \( m \) on a horizontal surface, with a rope attached to it. The rope is at an angle \( \theta \) above the horizontal. The forces acting on the crate include the tension in the rope \( F_T \).
**Questions & Solutions:**
**(a) Draw and label all of the forces acting on the crate.**
*Solution:*
The forces acting on the crate are:
- The gravitational force (\( F_g = mg \)) acting downwards.
- The normal force (\( F_N \)) acting perpendicular to the surface and upwards.
- The tension force (\( F_T \)) along the rope, which can be resolved into two components: horizontal (\( F_{T_x} = F_T \cos\theta \)) and vertical (\( F_{T_y} = F_T \sin\theta \)).
- The frictional force (\( F_f \)) acting opposite to the direction of motion, which is calculated as \( F_f = \mu F_N \).
**(b) Compute the normal force acting on the crate in terms of \( m \), \( F_T \), \( \theta \), and \( g \).**
*Solution:*
The normal force (\( F_N \)) can be found by balancing the vertical forces:
\[ F_N + F_T \sin \theta = mg \]
Thus,
\[ F_N = mg - F_T \sin \theta \]
**(c) Compute the acceleration of the crate in terms of \( m \), \( F_T \), \( \theta \), \( \mu \), and \( g \).**
*Solution:*
First, determine the net horizontal force (\( F_{net} \)) acting on the crate:
\[ F_{net} = F_T \cos \theta - F_f \]
\[ F_f = \mu F_N = \mu (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90b2ad7e-71be-4993-92e0-fe2030354ee3%2F650431e7-0063-468e-8ec7-e683074fb87f%2Fa9ag83i_processed.png&w=3840&q=75)
Transcribed Image Text:**1. Topic: Motion of a Crate Being Pulled Across a Horizontal Surface**
This exercise addresses the motion of a crate that is being pulled across a horizontal floor by a rope. In the diagram provided, the following variables are defined:
- \( m \): Mass of the crate
- \( \mu \): Coefficient of kinetic friction between the crate and the floor
- \( F_T \): Tension in the rope
- \( \theta \): Angle of the rope above the horizontal
**Diagram Explanation:**
The diagram shows a crate of mass \( m \) on a horizontal surface, with a rope attached to it. The rope is at an angle \( \theta \) above the horizontal. The forces acting on the crate include the tension in the rope \( F_T \).
**Questions & Solutions:**
**(a) Draw and label all of the forces acting on the crate.**
*Solution:*
The forces acting on the crate are:
- The gravitational force (\( F_g = mg \)) acting downwards.
- The normal force (\( F_N \)) acting perpendicular to the surface and upwards.
- The tension force (\( F_T \)) along the rope, which can be resolved into two components: horizontal (\( F_{T_x} = F_T \cos\theta \)) and vertical (\( F_{T_y} = F_T \sin\theta \)).
- The frictional force (\( F_f \)) acting opposite to the direction of motion, which is calculated as \( F_f = \mu F_N \).
**(b) Compute the normal force acting on the crate in terms of \( m \), \( F_T \), \( \theta \), and \( g \).**
*Solution:*
The normal force (\( F_N \)) can be found by balancing the vertical forces:
\[ F_N + F_T \sin \theta = mg \]
Thus,
\[ F_N = mg - F_T \sin \theta \]
**(c) Compute the acceleration of the crate in terms of \( m \), \( F_T \), \( \theta \), \( \mu \), and \( g \).**
*Solution:*
First, determine the net horizontal force (\( F_{net} \)) acting on the crate:
\[ F_{net} = F_T \cos \theta - F_f \]
\[ F_f = \mu F_N = \mu (
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