Statics - 2D equilibrium - particle problems - example
Summary
TLDRThis 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
- š§āš« The problem involves a 100 kg mass supported by three cables (AB, AC, and AD) with specific dimensions and angles.
- š AB forms a 50-degree angle with the horizontal, while AC and AD connect to the ceiling and the mass at different points.
- š 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.
- āļø A Freebody diagram is drawn, cutting through the cables to reveal forces acting on the system, including gravitational force (100 kg Ć 9.81 m/sĀ²).
- š§ 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.
- ā 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.
- š 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.
- āļø 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.
- š§® Algebraic manipulations lead to the equation FA = 1.245 FAC, and further solving yields numerical values for the forces in the cables.
- ā 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
š 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.
š 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.
š§® 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.
Mindmap
Keywords
š”Freebody diagram
š”Cable forces
š”Equilibrium
š”Force components
š”Trigonometry
š”Similar triangles
š”Newton's second law
š”Acceleration due to gravity
š”Right triangle
š”Vector
š”Sum of forces
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 thisĀ problem we have a math of a hundred kilogramsĀ Ā
which is supported by three tables we have aĀ B which is going up to the ceiling we have ACĀ Ā
which is going up to the ceiling and we have a DĀ which connects between this ring and 100-kilogramĀ Ā
master you can see the dimensions that we haveĀ over here we have a three meter drop from theĀ Ā
ceiling down to point a and we have a four metersĀ over from Point C to point a over here on a B weĀ Ā
don't have the dimensions but we do have thatĀ there's a 50 degree angle that a B makes fromĀ Ā
the horizontal so let's start off with drawingĀ a Freebody diagram and then I'll show you how toĀ Ā
solve it with our free body diagram let's imagineĀ that we have a pair of scissors we're going to cutĀ Ā
through all the cables that are supporting a it isĀ going to be our particle it's one point in spaceĀ Ā
so we're going to redraw a over here so here'sĀ our ring a where everything connects togetherĀ Ā
then when we cut through this cable we instead ofĀ put the force that's in the cable this was a massĀ Ā
of 100 kilograms and so we need to multiply thatĀ by 9.81 so that we can convert it into these soĀ Ā
we have this 100 kilograms times 9.81 for fa bĀ we cut through the card a B over here and so toĀ Ā
show the force that is in that cable we want toĀ label it something like F or T and then use theĀ Ā
the two points of a B of F a B or T a P we willĀ also ensure that it's a vector and so we'll goĀ Ā
ahead and put the vector symbol on top to showĀ that that's the vector quantity over here we cutĀ Ā
through a C and so to show the force it isn't aĀ see will label it again f40 and since it's an ACĀ Ā
FAC and then again with the arrow on top to showĀ that it's the vector because we have the distanceĀ Ā
information over there we can draw similarĀ triangle on this force so the four meters here wasĀ Ā
parallel to the horizontal parallel to the x-axisĀ so the four goes down on the bottom the threeĀ Ā
meters to mention here is parallel to the verticalĀ it's parallel to the y axis and so the three goesĀ Ā
on this vertical side over here when you takeĀ 3 squared plus 4 squared and then you take aĀ Ā
square root you get five it's three four fiveĀ right triangle so we've got that three four fiveĀ Ā
right train will now sketch in and it's going toĀ give us the direction information on FSC now let'sĀ Ā
go ahead and instead of our sum of the forcesĀ equations and extracting and we can start solvingĀ Ā
make sure when you're writing these equationsĀ that you do rights on the forces in the x equalsĀ Ā
zero net hooks the greater or the person of theĀ professor for professors grading the test kind ofĀ Ā
follow your work here so we have sum of forces inĀ the x-direction is equal to zero and we have the xĀ Ā
component from that baby so the magnitude of X a BĀ times the cosine of 50 degrees we're using cosineĀ Ā
because the X side is adjacent to that angle itĀ touches the angle of 50 degrees then over hereĀ Ā
with our similar triangles we're going to stillĀ use the magnitude of FAC and then we're goingĀ Ā
to multiply it by a ratio of one of the sidesĀ to the hypotenuse remember that hypothesis isĀ Ā
this longer side here of five so when we're doingĀ this this is going in the X direction we're goingĀ Ā
to pick this portion of the similar triangle thatĀ is in the X direction so we're going to pick theĀ Ā
four side there for side is on top divided byĀ the 5 sine which is our highpockets now let'sĀ Ā
go through and do the algebra real quick on thisĀ one and just a real quick look we've got cosineĀ Ā
of 50 degrees so just pathetic a calculator getĀ a decimal for that and then you have four-fifthsĀ Ā
converted to decimals your point eight over hereĀ both of these two terms are equal to zero so we'reĀ Ā
going to add this negative terms of both sides soĀ it's going to move this over to the other side ofĀ Ā
the equation so I have a point six four two eightĀ fa b is not equal to zero point eight FAC we'reĀ Ā
going to divide both sides of the equation by theĀ point six four two eight that gives us F a B isĀ Ā
one point two four five FSC now once you are someĀ two forces in the Y so here's our several forcesĀ Ā
in the Y equation we have the y component of F aĀ B it's going to be FA B sine of 50 degrees we'reĀ Ā
using sine because we're looking at the verticalĀ side here which is opposite to the table and IĀ Ā
would have FAC times the vertical side becauseĀ we're doing some forces in the wine which isĀ Ā
going to be the three side divided by the promisedĀ which is the five side then down here we have aĀ Ā
negative 981 Newton to do - 981 and that's allĀ equal to zero again when you're writing eitherĀ Ā
one of these equations make sure that you writeĀ the sum of forces use the direction and say thatĀ Ā
it's equal to zero it helps people follow yourĀ work now let's send check that the house overĀ Ā
here will get a solution so just like beforeĀ we'll take the sign and just go ahead and putĀ Ā
in a calculator convert it out hit a decimal doĀ the three fifths I broke that over to the decimalĀ Ā
we've got the negative 981 here so we're going toĀ add 981 to both sides so 7i 981 over to this sideĀ Ā
of the equation and then we have this expressionĀ this equation that we already developed from sumĀ Ā
of forces and X we'll go ahead and plug that inĀ so where we have FA B over here we're instead ofĀ Ā
going to plug in one point two four five fac-1 soĀ you can sit up and plug in there and then we'llĀ Ā
go ahead and make the multiplication with theĀ coefficients and then add that coefficient toĀ Ā
it that will give this one point five five forĀ FA CE is equal to nine eighty one business andĀ Ā
divide through both sides here by the one pointĀ five five four and we get FA CE is equal to sixĀ Ā
thirty one we'll take that 631 plug it back andĀ up here multiplied by the one point two four fiveĀ Ā
and we'll get the 786 Newton's therefore the FACĀ and we're done on this problem thanks for watching
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