Example Question - magnitude
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Softball Player Sliding Across Infield Dirt and Forces Involved
<p>When a softball player is sliding across the infield dirt, there are several forces acting upon the player. Assuming the player is moving to the right, the types of forces and their relative magnitudes would typically include:</p> - <p>Gravity: Acting downward with a magnitude proportional to the mass of the player. This is a constant force.</p> - <p>Normal force: Acting upward exerted by the ground on the player, equal in magnitude and opposite in direction to gravity when no vertical acceleration is present.</p> - <p>Friction: Acting to the left, opposite the direction of motion, with a magnitude dependent on the coefficient of friction between the player and the dirt and the normal force.</p> <p>The gravitational force and the normal force are usually equal in magnitude if there is no vertical acceleration. Friction is typically less than these forces for a sliding object unless the coefficient of friction is exceptionally high.</p>
Softball Player Sliding Forces Analysis
<p>The forces acting on the softball player during a head-first dive across the infield dirt while ignoring air resistance are:</p> <p>1. Gravity (weight), which acts downwards and is equal to the mass of the softball player times the acceleration due to gravity (mg).</p> <p>2. Normal force, which acts perpendicular to the contact surface and upwards. It is equal in magnitude and opposite in direction to the weight if the player is sliding on a level surface.</p> <p>3. Frictional force, which acts parallel to the contact surface and opposite to the direction of motion. The magnitude of the frictional force depends on the coefficient of friction between the player and the infield dirt and the normal force.</p> <p>The magnitudes of these forces are relative to each other: the normal force balances the weight in the absence of vertical acceleration, and the frictional force is less than the weight and normal force since it opposes the motion without stopping it immediately (assuming the player keeps sliding).</p>
Forces Acting on a Softball Player During a Head-First Dive
<p>The problem involves analyzing the forces acting on a softball player during a head-first slide. Since the image is not displayed, we will consider the typical forces acting in this scenario without air resistance:</p> <p>1. Gravitational force (weight) acting downwards. Denoted as \( F_g \) and is equal to \( m \times g \), where \( m \) is the mass of the player and \( g \) is the acceleration due to gravity.</p> <p>2. Normal force acting upwards from the ground, opposite to the gravitational force. Denoted as \( F_N \) and typically equal in magnitude to \( F_g \) in a scenario without vertical acceleration. Thus, \( F_N = F_g \).</p> <p>3. Frictional force acting opposite to the direction of the slide, which is to the left if the player is sliding to the right. The magnitude of the frictional force (\( F_f \)) depends on the coefficient of friction (\( \mu \)) between the player and the ground and the normal force. So, \( F_f = \mu \times F_N \).</p> <p>There are no other forces acting in this scenario since we are ignoring air resistance. The relative magnitude of these forces will be (\( F_N = F_g \)) greater than \( F_f \), assuming a realistic coefficient of friction.</p>
Determining the Resultant Force and its Direction
<p>Given:</p> <p>\(\vec{F_1} = 800 \text{ N} \ (\text{upwards along the y-axis})\)</p> <p>\(\vec{F_2} = 600 \text{ N} \ (\text{30 degrees below the -x-axis})\)</p> <p>To solve for the resultant force \(\vec{R}\), break \(\vec{F_2}\) into its x and y components.</p> <p>\(F_{2x} = 600 \cos(30^\circ)\)</p> <p>\(F_{2y} = -600 \sin(30^\circ)\) (The y-component is negative because it is downward.)</p> <p>Calculate \(F_{2x}\) and \(F_{2y}\):</p> <p>\(F_{2x} = 600 \times \frac{\sqrt{3}}{2} = 300\sqrt{3}\) N</p> <p>\(F_{2y} = -600 \times \frac{1}{2} = -300\) N</p> <p>The resultant force \(\vec{R}\) in the x and y components is:</p> <p>\(R_x = F_{2x}\) because there is no x-component for \(\vec{F_1}\).</p> <p>\(R_y = F_{1} + F_{2y}\)</p> <p>Calculate \(R_x\) and \(R_y\):</p> <p>\(R_x = 300\sqrt{3}\) N</p> <p>\(R_y = 800 - 300 = 500\) N</p> <p>Now calculate the magnitude of the resultant vector \(|\vec{R}|\):</p> <p>\(|\vec{R}| = \sqrt{R_x^2 + R_y^2}\)</p> <p>\(|\vec{R}| = \sqrt{(300\sqrt{3})^2 + (500)^2}\)</p> <p>\(|\vec{R}| = \sqrt{(900 \times 3) + (250000)}\)</p> <p>\(|\vec{R}| = \sqrt{2700 + 250000}\)</p> <p>\(|\vec{R}| = \sqrt{252700}\)</p> <p>\(|\vec{R}| \approx 502.7\) N</p> <p>To find the direction θ measured counterclockwise from the positive x-axis:</p> <p>\(\tan(\theta) = \frac{R_y}{R_x}\)</p> <p>\(\theta = \arctan\left(\frac{500}{300\sqrt{3}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{300 \times \frac{\sqrt{3}}{2}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{500}{150\sqrt{3}}\right)\)</p> <p>\(\theta \approx \arctan\left(\frac{500}{259.81}\right)\)</p> <p>\(\theta \approx \arctan(1.925)\)</p> <p>\(\theta \approx 62.5^\circ\) (above the negative x-axis)</p> <p>Finally, from the positive x-axis, add 180 degrees because it is on the left half of the coordinate system:</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + \theta\)</p> <p>\(\theta_{\text{from positive x-axis}} = 180^\circ + 62.5^\circ\)</p> <p>\(\theta_{\text{from positive x-axis}} = 242.5^\circ\)</p>
Determining the Resultant Force and its Direction
<p>To find the magnitude of the resultant force, we can break the forces into their x and y components. For the 600 N force:</p> <p>\[ F_{x1} = 600 \cos(30^\circ) \]</p> <p>\[ F_{y1} = 600 \sin(30^\circ) \]</p> <p>For the 800 N force, which is purely in the y-direction:</p> <p>\[ F_{x2} = 0 \]</p> <p>\[ F_{y2} = 800 \]</p> <p>Now, add the x-components and y-components of the forces to find the resultant:</p> <p>\[ R_x = F_{x1} + F_{x2} \]</p> <p>\[ R_y = F_{y1} + F_{y2} \]</p> <p>Substitute the values to find \( R_x \) and \( R_y \):</p> <p>\[ R_x = 600 \cos(30^\circ) \]</p> <p>\[ R_y = 600 \sin(30^\circ) + 800 \]</p> <p>Use the Pythagorean theorem to find the magnitude of the resultant force (R):</p> <p>\[ R = \sqrt{R_x^2 + R_y^2} \]</p> <p>Calculate the angle (\( \theta \)) from the positive x-axis:</p> <p>\[ \theta = \arctan\left(\frac{R_y}{R_x}\right) \]</p> <p>Solve for \( R \) and \( \theta \) to get the magnitude and direction of the resultant force.</p>
Determination of Resultant Force Magnitude and Direction
For Prob. F2-1: <p>\(R_{x} = 6 \cos(45^\circ)N - 2N\)</p> <p>\(R_{x} = 4.2426N - 2N\)</p> <p>\(R_{x} = 2.2426N\)</p> <p>\(R_{y} = 6 \sin(45^\circ)N\)</p> <p>\(R_{y} = 4.2426N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{2.2426^2 + 4.2426^2}N\)</p> <p>\(R = \sqrt{5.0291 + 18.0000}N\)</p> <p>\(R = \sqrt{23.0291}N\)</p> <p>\(R \approx 4.8001N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{4.2426}{2.2426}\right)\)</p> <p>\(\theta \approx 62.1021^\circ\) clockwise from the +x-axis</p> For Prob. F2-2: Isolated from the image and omitted based on the provided instructions. For Prob. F2-3: <p>\(R_{x} = 800 \cos(30^\circ)N - 600 \cos(45^\circ)N\)</p> <p>\(R_{x} = 692.82N - 424.26N\)</p> <p>\(R_{x} = 268.56N\)</p> <p>\(R_{y} = 800 \sin(30^\circ)N + 600 \sin(45^\circ)N\)</p> <p>\(R_{y} = 400N + 424.26N\)</p> <p>\(R_{y} = 824.26N\)</p> <p>\(R = \sqrt{R_{x}^2 + R_{y}^2}\)</p> <p>\(R = \sqrt{268.56^2 + 824.26^2}N\)</p> <p>\(R = \sqrt{72105.59 + 679186.67}N\)</p> <p>\(R = \sqrt{751292.27}N\)</p> <p>\(R \approx 866.77N\)</p> <p>\(\theta = \arctan\left(\frac{R_{y}}{R_{x}}\right)\)</p> <p>\(\theta = \arctan\left(\frac{824.26}{268.56}\right)\)</p> <p>\(\theta \approx 71.5651^\circ\) counterclockwise from the +x-axis</p>
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