Equilibrium of Concurrent Forces Group D PDF

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GROUP D: MARIANA RUIZ SOFIA PILATO CRISTIAN PEREZ

Equilibrium of Concurrent Forces OBJECTIVES: ● ●

To study the equilibrium of a body under the action of concurrent forces. To verify Newton’s First Law.

MATERIALS: ● ● ● ● ● ● ●

Force Table Centering ring Small rod to hold the ring 4 pieces of string 4 pulleys 4 metal hooks Slotted mass set (2 of 50g and 4 of 100g)

THEORY. Force is one of the fundamental concepts upon which the subject of mechanics is built. A force is the action of one body on another body which changes or tends to change the state of motion of the body on which the action is taking place. The idea of force, then, implies the mutual actions of two bodies, since one body cannot exert a force on another body unless the second object offers a resistance to the first. A force, therefore, never exists alone. Forces always come in pairs. If a body is acted on by one force only, a change of motion of the body will always take place: but if the body is acted on by two or more forces, it may be held at rest. Concurrent forces are those whose lines of action all pass through the same point. 

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 Newton’s First Law states, “Every object continues to be at rest or to move with constant velocity in straight line unless a net force acts on it.”  If ∑ F = 0 V = 0 m/s

(the object is at rest)

V = constant (The object is moving with constant velocity in straight line) Force is a vector quantity. It has magnitude and direction. We can represent a force through its components.  

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    ∑ F = 0 then ∑ Fx = 0 ∑ Fy = 0. It is known as the equilibrium condition for translational motion.  The force table below is used to verify Newton’s First Law in this experiment. Forces (weights) of any desire magnitude may be attached to cords, which run over pulleys that can be set at any desire position (angle) around a 3600 circle. The other end of each cord is attached to a small ring, which is held in the center of the force table by a pin until equilibrium is attained.

PROCEDURE: ● ● ● ● ● ● ● ● ● ● ● ● ●

Set up the force table with two pulleys and two equal weights hanging from the hooks. Move the pulleys until the ring is set in the center of the force table. Measure the angle for each cord (which corresponds to the force due to the weight hanging). Find for this force the X component (Fx) and the Y component ( Fy). Add one more pulley (now you have three) and hang three different weights from the hooks. Move the pulleys until setting the ring in the center of the force table. Measure the angle for each cord (which corresponds to the force due to the weight hanging). Find for this force the X component (Fx) and the Y component (Fy). Add one more pulley (now you have four) and hang four different weights from the hooks. Move the pulleys until the ring is set in the center of the force table. Measure the angle for each cord (which corresponds to the force due to the weight hanging). Find for this force the X component (Fx) and the Y component (Fy). Verify the equilibrium condition for translational motion

DATA Force (N) Mass(kg)

Angle( º )

Fsinθ(N)

Fx (N)

Fy (N)

cos θ

sin θ

Fcosθ(N)

Mass*Gravity 0.050

0

0.49

0

0.49

0.49

0

0.050

180

0.49

0

-0.49

-0.49

0

∑ Fx =0

∑ Fy =0

Net (N)

Force

=0 (At Rest)

∑F=0

0.050

261

0.49

-0.48

-0.08

-0.08

-0.48

0.100

277

0.98

-0.97

0.12

0.12

-0.97

0.150

90

1.47

1.47

0

0

1.47

∑Fx=0.04

∑Fy=0.02

0.050

90

0.49

0.49

0

0

0.49

0.100

250

0.98

-0.92

-0.34

-0.34

-0.92

0.150

180

1.47

0

-1.47

-1.47

0

0.200

10

1.96

0.34

1.93

1.93

0.34

∑Fx =0.12

∑Fy=-0.09

∑ F=0.06

∑ F=0.03

CONCLUSIONS: 1) We accomplished objective 1 by demonstrating  that  even  under  different conditions if the  concurrent  forcesare equaland opposite,thenetforceis0 and they will reach equilibrium. This isbecause of Newton’s firstlawthatsays that if forces cancel each other out the forces are at rest.Throughout the experiment we  noticed that when calculating the  net force with the data collected it was very close to zero. 2) In this experiment we put to test Newton’s first law by using the  force  table. Newton’s firstlawstatesthat “Everyobjectcontinuestobeat restor to move with constant velocity in a straightlineunlessanetforceactsonit.”  We test this law out because we are applying several forces that will cancel eachotheroutwhen putatacertain angle around the 360 degree force table.  As a result, the net force is in equilibrium because the forces are at rest, making Newton’s first law correct. Errors: This experiment consisted of various trials with different conditions. While reporting the data manyotherthingswere not taken into account, such as the weight of the string and the pulley. By missingthesemeasurementswe can say that the mass hanging fromthestringis not completelyaccuratewhen writteninthe data.Anotherlimitation in conducting the experimentaccurately would be the use of the materials such as the string. Somestingswerelonger than others, this could affect the results. This is an experiment that is conducted withhumanerrors.Thedatacollectedwillnotalwaysbe correctand  that's why the results for Net Force are not exactly zero....