{"id":1274,"date":"2021-06-29T20:19:35","date_gmt":"2021-06-29T20:19:35","guid":{"rendered":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/?p=1274"},"modified":"2021-12-30T19:07:47","modified_gmt":"2021-12-30T19:07:47","slug":"sections-9-5-v2","status":"publish","type":"post","link":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2021\/06\/29\/sections-9-5-v2\/","title":{"rendered":"Chapter 9: Force"},"content":{"rendered":"\n

9.5 Mechanical equilibrium<\/h2>\n\n\n\n

An object (or group of objects) is said to be in mechanical equilibrium<\/em> when the net force on it is zero. You could also describe this in terms of an object\u2019s motion, considering Newton\u2019s second law: an object is in mechanical equilibrium when its momentum is not changing, or in other words, when is not accelerating.<\/p>\n\n\n\n

This describes many things that we encounter in the world around us. For example, take a look at any building\u2014if any piece of a building has an acceleration other than zero, you do not want to be in or near that building!<\/p>\n\n\n\n

Example 9.4<\/h4>\n\n\n\n

You and a friend are dragging a 75 kg box of potatoes across a floor at a constant speed. You are pushing with a force of 100 N. Your friend is pulling on a rope with a force of 200 N What is the force of kinetic friction that the floor exerts on the box if your friend is pulling up at a 25\u00b0 angle? What is the normal force the ground exerts on the box?<\/p>\n\n\n\n

FBD\u2014Box<\/h5>\n\n\n\n
\"A
Free body diagram for the box of potatoes<\/figcaption><\/figure><\/div>\n\n\n\n
Newton’s second law:<\/h5>\n\n\n

\\[
\n \\begin{align*}
\n F_\\textit{net,x} &= ma_x & F_\\textit{net,y} &= ma_y \\\\
\n F^T_x + F^\\textit{push} – F^f &= m(0) & F^N + F^T_y – F^G &= m(0) \\\\
\n F^T\\cos\\theta + F^\\textit{push} – F^f &= 0 & F^N + F^T\\sin\\theta – mg &= 0 \\\\
\n F^f &= F^T\\cos\\theta – F^\\textit{push} & F^N &= mg – F^T\\sin\\theta \\\\
\n &= (200\\ \\textrm{N})\\cos(25^\\circ) – 100\\ \\textrm{N} & &= (75\\ \\textrm{kg})\\left(9.81\\ \\textrm{m\/s}^2\\right) – (200\\ \\textrm{N})\\sin(25^\\circ) \\\\
\n &= 81.3\\ \\textrm{N} & &= 651\\ \\textrm{N}
\n \\end{align*}
\n\\]<\/p>\n\n\n\n

Example 9.5<\/h4>\n\n\n\n

A bag full of books \\(\\left(m = 5.3\\ \\textrm{kg}\\right)\\) is resting on a table. You attach a spring, which has a spring constant of 150 N\/m to the bag, and pull horizontally. The spring is stretched 2.5 cm just before<\/em> the bag begins to move. What is the force of static friction that the table exerts on the bag?<\/p>\n\n\n\n

FBD\u2014Bag<\/h5>\n\n\n\n
\"A
Free body diagram for the bag of books<\/figcaption><\/figure><\/div>\n\n\n\n
Newton’s second law<\/h5>\n\n\n\n

Just before<\/em> the bag begins to move, its acceleration is zero.<\/p>\n\n\n

\\[
\n \\begin{align*}
\n F_\\textit{net,x} &= ma_x \\\\
\n F^\\textit{sp} – F^f &= m(0) \\\\
\n kx – F^f &= 0 \\\\
\n F^f &= kx \\\\
\n &= (150\\ \\textrm{N\/m})(2.5 \\times 10^{-2}\\ \\textrm{m}) \\\\
\n &= 3.75\\ \\textrm{N}
\n \\end{align*}
\n\\]<\/p>\n\n\n\n

Example 9.6<\/h4>\n\n\n\n

A flower basket is hanging from two wires. One wire is attached at a right angle to a wall, and the other is attached to the ceiling at an angle of 30\u00b0 to the horizontal. The mass of the basket is 6 kg. What is the tension in each wire?<\/p>\n\n\n\n

FBD\u2014Flower basket<\/h5>\n\n\n\n
\"A
Free body diagram for the flower basket<\/figcaption><\/figure><\/div>\n\n\n\n

In the previous problems, we could find out everything we wanted by just using the \\(x\\) or \\(y\\) components of the forces. For this problem, we will need to solve a system of two equations to find the tension in each wire.<\/p>\n\n\n\n

Newton’s second law<\/h5>\n\n\n\n\n\n

Now, we’ll find \\(F^T_2\\) using the \\(y\\) components of the forces, and do a substitution to find \\(F^T_1\\).<\/p>\n\n\n

\\[
\n\\begin{align*}
\n F^T_1 &= F^T_2\\cos\\theta & F^T_2 &= \\frac{mg}{\\sin\\theta} \\\\
\n &= \\left(\\frac{mg}{\\sin\\theta}\\right)\\cos\\theta & &= \\frac{(6\\ \\textrm{kg})\\left(9.81\\ \\textrm{m\/s}^2\\right)}{\\sin(30^\\circ)} \\\\
\n &= \\left(\\frac{(6\\ \\textrm{kg})\\left(9.81\\ \\textrm{m\/s}^2\\right)}{\\sin(30^\\circ)}\\right)\\cos(30^\\circ) & &= 117.7\\ \\textrm{N} \\\\
\n &= 101.9\\ \\textrm{N}
\n\\end{align*}
\n\\]<\/p>\n\n\n\n

You’ll notice I algebraically substituted \\(\\frac{mg}{\\sin\\theta}\\) for \\(F^T_2\\), rather than simply using the value 117.7 N. I do this for two reasons: you can often discover interesting relationships between the variables by focusing on the physical quantities rather than specific values; and you avoid errors that arise from rounding.<\/p>\n\n\n\n

Practice 9.6<\/h4>\n\n\n\n\n