Statics - 2D equilibrium - particle problems - example

Summary

This tutorial explains how to solve a physics problem involving a 100 kg mass supported by three cables. The instructor draws a free body diagram, labels forces, and uses trigonometry to set up equations for forces in the x and y directions. The tutorial demonstrates solving for cable tensions using the Pythagorean theorem and algebraic manipulation, resulting in the forces in cables AB and AC.

Takeaways

• 🧑‍🏫 The problem involves a 100 kg mass supported by three cables (AB, AC, and AD) with specific dimensions and angles.
• 📏 AB forms a 50-degree angle with the horizontal, while AC and AD connect to the ceiling and the mass at different points.
• 📐 The vertical distance from point A to the ceiling is 3 meters, and the horizontal distance from point C to point A is 4 meters, forming a 3-4-5 right triangle.
• ✂️ A Freebody diagram is drawn, cutting through the cables to reveal forces acting on the system, including gravitational force (100 kg × 9.81 m/s²).
• 🔧 Force vectors in the cables are labeled as FA, FB, and FAC to denote forces in different directions, with vector symbols used to represent magnitude and direction.
• ➗ The force in AB is decomposed into horizontal (cosine of 50 degrees) and vertical (sine of 50 degrees) components to analyze the forces acting on the system.
• 📊 A similar triangle approach is applied to AC using the known dimensions (3 meters and 4 meters), calculating the force in the x and y directions.
• ⚖️ The equations of equilibrium for both the x and y directions (sum of forces equals zero) are used to solve for unknown forces in the cables.
• 🧮 Algebraic manipulations lead to the equation FA = 1.245 FAC, and further solving yields numerical values for the forces in the cables.
• ✅ The final calculated forces are FA = 786 Newtons and FAC = 631 Newtons, completing the solution for the problem.

Q & A

What is the total weight of the mass in Newtons?

The mass is 100 kilograms, and the weight is calculated by multiplying the mass by the acceleration due to gravity (9.81 m/s^2), resulting in 100 kg * 9.81 m/s^2 = 981 Newtons.

What is the purpose of drawing a Free Body Diagram (FBD) in this scenario?

A Free Body Diagram is used to visualize all the forces acting on an object, in this case, the ring supporting the mass. It helps in solving the problem by breaking down the forces into components that can be mathematically analyzed.

What are the dimensions given for the problem?

The problem provides a 3-meter vertical drop from the ceiling to point A, and a 4-meter horizontal distance from point C to point A.

What is the angle given for cable AB?

Cable AB makes a 50-degree angle with the horizontal.

How is the force in cable AB represented in the FBD?

In the FBD, the force in cable AB is represented by a vector labeled as FAB or TAB, indicating that it is a vector quantity.

What is the significance of the 3-4-5 right triangle mentioned in the script?

The 3-4-5 right triangle is used to establish the relationship between the horizontal and vertical components of the forces in the cables. It helps in determining the direction of the forces acting on the ring.

How are the forces in the x-direction calculated?

The forces in the x-direction are calculated using the horizontal component of FAB (FAB * cos(50 degrees)) and the horizontal component of FAC (FAC * (4/5)), where 4 is the horizontal side and 5 is the hypotenuse of the 3-4-5 triangle.

What is the sum of forces equation in the x-direction?

The sum of forces in the x-direction is set to zero: FAB * cos(50 degrees) + FAC * (4/5) = 0.

How are the forces in the y-direction calculated?

The forces in the y-direction are calculated using the vertical component of FAB (FAB * sin(50 degrees)) and the vertical component of FAC (FAC * (3/5)), where 3 is the vertical side and 5 is the hypotenuse of the 3-4-5 triangle.

What is the sum of forces equation in the y-direction?

The sum of forces in the y-direction is set to zero: FAB * sin(50 degrees) + FAC * (3/5) - 981 = 0.

How are the forces FAB and FAC solved for?

The forces FAB and FAC are solved for by setting up the sum of forces equations in both the x and y directions, and then solving the system of equations.

Outlines

00:00

📐 Mechanics Problem Introduction

The speaker introduces a tutorial on mechanics, focusing on a problem involving a 100-kilogram mass supported by three tables. The setup includes a vertical table (B), another table (AC) also vertical, and a third table (D) connecting a ring to the mass. The dimensions provided are a 3-meter drop from the ceiling to point A and a 4-meter horizontal distance from point C to point A. The angle of table AB relative to the horizontal is 50 degrees. The tutorial will involve drawing a free body diagram to solve the problem, starting with cutting through the cables supporting the mass to represent the forces in the cables as vectors.

05:04

🔍 Free Body Diagram and Force Calculation

The tutorial continues with the creation of a free body diagram for the mass. The forces in the cables are represented as vectors, with the force in cable AB labeled as F_AB or T_AB and the force in cable AC as F_AC. Using trigonometry and the Pythagorean theorem, the forces are calculated by considering the horizontal and vertical components. The horizontal force is calculated using the cosine of the 50-degree angle, and the vertical force is calculated using the sine of the same angle. The tutorial emphasizes the importance of writing equations for the sum of forces in the correct direction and equaling them to zero. The algebraic manipulation is demonstrated to solve for the forces in the cables.

🧮 Solving for Forces in Y-Direction

The final part of the tutorial involves solving for the forces in the Y-direction. The speaker uses the previously calculated forces to find the vertical component of F_AB and the vertical side of the similar triangle to find F_AC. The equation is set up with the sum of forces in the Y-direction equal to zero, including the weight of the mass. The algebra is performed to solve for F_AC, and the result is found to be 631 Newtons. The tutorial concludes with the final force values and a thank you to the viewers.

Keywords

💡Freebody diagram

A freebody diagram is a graphical representation of all the forces acting on an object, isolated from its surroundings. In the video, the instructor uses a freebody diagram to analyze the forces acting on a 100-kilogram mass supported by three cables. The diagram helps to visualize and solve the problem by breaking down the forces into their horizontal and vertical components.

💡Cable forces

Cable forces refer to the tension forces within the cables that are supporting the weight. In the script, the instructor calculates the tension in cables AB and AC by considering the weight of the mass and the angles involved. These forces are crucial for determining the equilibrium of the system.

💡Equilibrium

Equilibrium in physics means that the net force acting on an object is zero, and the object is either at rest or moving at a constant velocity. The video's main theme revolves around finding the equilibrium state of the 100-kilogram mass by ensuring the sum of forces in both the x and y directions equals zero.

💡Force components

Force components are the individual parts of a force vector that are aligned with the coordinate axes. In the script, the instructor decomposes the forces in cables AB and AC into horizontal (x-component) and vertical (y-component) parts to apply them in the equilibrium equations.

💡Trigonometry

Trigonometry is used to relate the angles of a triangle to the ratios of its sides. The instructor applies trigonometry to find the horizontal and vertical components of the cable forces by using cosine and sine functions, which correspond to the adjacent and opposite sides of the angles formed by the cables.

💡Similar triangles

Similar triangles are triangles that have the same shape but not necessarily the same size. In the video, the instructor uses the concept of similar triangles to relate the forces in the cables to the dimensions of the physical setup, allowing for the calculation of the forces based on the known lengths.

💡Newton's second law

Newton's second law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. Although not explicitly mentioned, the law is implicitly used when the instructor converts the weight of the mass into force by multiplying the mass by the acceleration due to gravity (9.81 m/s^2).

💡Acceleration due to gravity

The acceleration due to gravity is a constant (9.81 m/s^2) that represents the rate at which objects accelerate towards the Earth. In the script, it is used to convert the weight of the mass from a mass unit (kilograms) to a force unit (Newtons).

💡Right triangle

A right triangle is a triangle with one angle measuring 90 degrees. The instructor uses the properties of a right triangle (specifically the 3-4-5 Pythagorean triple) to determine the magnitude of the forces in the cables based on the given lengths and angles.

💡Vector

A vector is a quantity that has both magnitude and direction. In the video, forces are treated as vectors, and the instructor uses vector notation (an arrow above the force label) to indicate that the forces have both magnitude and direction, which is essential for solving the problem.

💡Sum of forces

The sum of forces refers to the algebraic addition of all the forces acting on an object. The instructor sets up equations for the sum of forces in the x and y directions to zero, which is a requirement for the object to be in equilibrium.

Highlights

Introduction of the problem: 100 kg mass supported by three cables.

Description of the cable arrangement with a 50-degree angle for cable AB.

Explanation of Freebody Diagram for the ring at point A, cutting through all supporting cables.

Multiplying 100 kg mass by 9.81 to convert to Newtons for force calculations.

Introduction of vector notation for the forces in cables AB, AC, and AD.

Using similar triangles to derive the directional components of the forces.

Details on calculating the right triangle dimensions: 3 meters drop and 4 meters horizontal distance.

Summing forces in the X-direction and solving for forces in cables AB and AC using trigonometric relationships.

Cosine of 50 degrees used to resolve forces in the X-direction.

Solving the equation for force in cable AB, resulting in a ratio between FAB and FAC.

Summing forces in the Y-direction and using sine functions to resolve forces.

Including gravitational force (981 N) in the Y-direction equations.

Substituting the earlier ratio into the Y-direction equation to solve for FAC.

Final solution for the forces: FAB = 786 N and FAC = 631 N.

Completion of the problem-solving process, summarizing the key results for the forces in cables.

Transcripts

hi I'm welcome to the status tutorials in this  problem we have a math of a hundred kilograms  

which is supported by three tables we have a  B which is going up to the ceiling we have AC  

which is going up to the ceiling and we have a D  which connects between this ring and 100-kilogram  

master you can see the dimensions that we have  over here we have a three meter drop from the  

ceiling down to point a and we have a four meters  over from Point C to point a over here on a B we  

don't have the dimensions but we do have that  there's a 50 degree angle that a B makes from  

the horizontal so let's start off with drawing  a Freebody diagram and then I'll show you how to  

solve it with our free body diagram let's imagine  that we have a pair of scissors we're going to cut  

through all the cables that are supporting a it is  going to be our particle it's one point in space  

so we're going to redraw a over here so here's  our ring a where everything connects together  

then when we cut through this cable we instead of  put the force that's in the cable this was a mass  

of 100 kilograms and so we need to multiply that  by 9.81 so that we can convert it into these so  

we have this 100 kilograms times 9.81 for fa b  we cut through the card a B over here and so to  

show the force that is in that cable we want to  label it something like F or T and then use the  

the two points of a B of F a B or T a P we will  also ensure that it's a vector and so we'll go  

ahead and put the vector symbol on top to show  that that's the vector quantity over here we cut  

through a C and so to show the force it isn't a  see will label it again f40 and since it's an AC  

FAC and then again with the arrow on top to show  that it's the vector because we have the distance  

information over there we can draw similar  triangle on this force so the four meters here was  

parallel to the horizontal parallel to the x-axis  so the four goes down on the bottom the three  

meters to mention here is parallel to the vertical  it's parallel to the y axis and so the three goes  

on this vertical side over here when you take  3 squared plus 4 squared and then you take a  

square root you get five it's three four five  right triangle so we've got that three four five  

right train will now sketch in and it's going to  give us the direction information on FSC now let's  

go ahead and instead of our sum of the forces  equations and extracting and we can start solving  

make sure when you're writing these equations  that you do rights on the forces in the x equals  

zero net hooks the greater or the person of the  professor for professors grading the test kind of  

follow your work here so we have sum of forces in  the x-direction is equal to zero and we have the x  

component from that baby so the magnitude of X a B  times the cosine of 50 degrees we're using cosine  

because the X side is adjacent to that angle it  touches the angle of 50 degrees then over here  

with our similar triangles we're going to still  use the magnitude of FAC and then we're going  

to multiply it by a ratio of one of the sides  to the hypotenuse remember that hypothesis is  

this longer side here of five so when we're doing  this this is going in the X direction we're going  

to pick this portion of the similar triangle that  is in the X direction so we're going to pick the  

four side there for side is on top divided by  the 5 sine which is our highpockets now let's  

go through and do the algebra real quick on this  one and just a real quick look we've got cosine  

of 50 degrees so just pathetic a calculator get  a decimal for that and then you have four-fifths  

converted to decimals your point eight over here  both of these two terms are equal to zero so we're  

going to add this negative terms of both sides so  it's going to move this over to the other side of  

the equation so I have a point six four two eight  fa b is not equal to zero point eight FAC we're  

going to divide both sides of the equation by the  point six four two eight that gives us F a B is  

one point two four five FSC now once you are some  two forces in the Y so here's our several forces  

in the Y equation we have the y component of F a  B it's going to be FA B sine of 50 degrees we're  

using sine because we're looking at the vertical  side here which is opposite to the table and I  

would have FAC times the vertical side because  we're doing some forces in the wine which is  

going to be the three side divided by the promised  which is the five side then down here we have a  

negative 981 Newton to do - 981 and that's all  equal to zero again when you're writing either  

one of these equations make sure that you write  the sum of forces use the direction and say that  

it's equal to zero it helps people follow your  work now let's send check that the house over  

here will get a solution so just like before  we'll take the sign and just go ahead and put  

in a calculator convert it out hit a decimal do  the three fifths I broke that over to the decimal  

we've got the negative 981 here so we're going to  add 981 to both sides so 7i 981 over to this side  

of the equation and then we have this expression  this equation that we already developed from sum  

of forces and X we'll go ahead and plug that in  so where we have FA B over here we're instead of  

going to plug in one point two four five fac-1 so  you can sit up and plug in there and then we'll  

go ahead and make the multiplication with the  coefficients and then add that coefficient to  

it that will give this one point five five for  FA CE is equal to nine eighty one business and  

divide through both sides here by the one point  five five four and we get FA CE is equal to six  

thirty one we'll take that 631 plug it back and  up here multiplied by the one point two four five  

and we'll get the 786 Newton's therefore the FAC  and we're done on this problem thanks for watching