To install StudyMoose App tap and then “Add to Home Screen”
Save to my list
Remove from my list
Static equilibrium of forces was investigated through the use of different weights attached to cords which were connected to a central ring, while pulleys supported them. This assembly facilitated the force band system to demonstrate that equilibrium will be attaining regardless of disturbances. However, due to errors in the experiment, the sum of the x and y component did equate to zero as predicted. The graphical solution of the experiment yield a polygon that is completed indicating that all the forces are in equilibrium while the analytical solution indicates a resultant force of 0.088N ± 0.181.
A particle is said to be in equilibrium when the vector sum of the external forces acting on a point is zero.
The analysis of particles in equilibrium is based on Newton’s 1st law of motion, which states that every object in a state of uniform motion tends to remain in that state unless acted upon by an external unbalanced force. When a body remains in a uniform state of motion, the sum of all the forces acting on it is zero, and it is in equilibrium.
Coplanar force systems have all the forces acting in one plane.
Verified writer
Proficient in: Physics
4.7 (346)
“ This writer never make an mistake for me always deliver long before due date. Am telling you man this writer is absolutely the best. ”
They may be concurrent, parallel, non-concurrent, or non-parallel. Almost any system of known forces can be resolved into a single force called a resultant force. The resultant is a representative force that has the same effect on the body as the group of forces it replaces. It can be determined both graphically and algebraically. The graphical methods are parallelogram and triangular, while algebraically, the resultant is the root of the summation square of the various forces along their individual axis.
It is important to note that for any given system of forces, there is only one resultant.
Forces can act upon a rigid body and also a particle in the same way. A particle has negligible dimensions and undergoes only translational motion, while a rigid body is one of definite shape and size. A rigid body will undergo negligible change when a force is applied, and it undergoes both translational and rotational motion. However, both will be in equilibrium if the summation of the forces acting on them is equal to zero.
ΣF = 0 ; ΣFx i + ΣFy j = 0
For this vector equation to be satisfied, the forces in x and y respectively must be equal to zero. Hence;
ΣFx = 0 , ΣFy = 0
When conducting an analysis on a rigid body, the first step is to do a free body diagram on the system. The steps in doing this are as follows:
Apparatus:
| Force (N) | Angle (°) (with respect to horizontal) | X-component (N) | Y-component (N) | |
|---|---|---|---|---|
| F1 | 1.77 | 37° | -1.41 | 1.06 |
| F2 | 1.08 | 28° | -0.953 | -0.507 |
| F3 | 1.08 | 90° | 0 | -1.08 |
| F4 | 1.18 | 30° | 1.02 | -0.590 |
| F5 | 1.77 | 37° | 1.41 | 1.06 |
ΣFx= -1.41 – 0.953 + 0 + 1.02 + 1.41= 0.067 N
ΣFy= 1.06 - 0.507 - 1.08 - 0.590 + 1.06 = -0.057 N
R= 0.0880 N
| Force (N) | Angle (°) (with respect to horizontal) | Cos (Angle) | X-component (N) | Sin (Angle) | Y-component (N) |
|---|---|---|---|---|---|
| F1 | 1.77 | 37° | 0.799 ± 0.025 | -1.41 ± 0.044 | 0.602 ± 0.025 |
| F2 | 1.08 | 28° | 0.882 ± 0.025 | -0.953 ± 0.027 | 0.470 ± 0.025 |
| F3 | 1.08 | 90° | 0 ± 0.025 | 0 | 1 ± 0.025 |
| F4 | 1.18 | 30° | 0.866 ± 0.025 | 1.02 ± 0.029 | 0.5 ± 0.025 |
| F5 | 1.77 | 37° | 0.799 ± 0.025 | 1.41 ± 0.044 | 0.602 ± 0.025 |
Resultant Uncertainty:
ΔR = √((ΔFx)^2 + (ΔFy)^2)
ΔR = √((0.044)^2 + (0.044)^2) = 0.0624 N
The results obtained in this experiment showed that errors were associated with the values. The system was in equilibrium as when it was disturbed it returned to its original position. Therefore, the errors may have been caused by the experimenter, the apparatus, and external environmental conditions, which gave the resultant 0.0880 N instead of zero (0).
Friction in the pulley could have hindered the smooth motion of the cords while the attached masses tried to maintain equilibrium, thus resulting in the inaccuracy in the angles. Parallax error is a major cause factor for the inaccuracy of the results obtained. The points that were placed on the drawing paper behind each cord may not have been exactly behind the individual cord as there was a significant gap between the paper and the cord. A third form of error that could have caused inaccuracy is environmental conditions. During the experiment, the force board was subjected to interference, as the experimenter caused some amount of vibration which shifts the board.
Regardless of the many precautionary measures made, the errors were unavoidable; however, the experiment was conducted in an accepted manner as the resultant was close with a small degree of uncertainty.
Concurrent forces intersect at one common point, and the resultant of these forces in the system acting on a body at rest equates to zero. However, the experimental values obtained for the resultant force are (FR = 0.0880 ± 0.181 N), which was calculated from the sum of Forces in the X direction and Y direction respectively.
It is recommended for this experiment to be more effective in future investigations and for confirmation of theories, that pulleys be situated so that the distance between the cords and the board are closer. This will reduce the errors that will be likely caused by parallax and vibration. A smoother force board can be used to reduce any negative impact that it can have on the experiment. I would also recommend repeating the experiment several times each time disturbing the system and ensuring that the cords line up exactly with the lines of action before removing the paper from the force board.
Force 1
Summation of Forces (in equilibrium): ΣFx= 0 ; ΣFy= 0 ; FR = √((ΣFx)^2 + (ΣFy)^2)
+→F1x = 1.77 N x Cos 37⁰ = -1.41 N
+↑F1y = 1.77 N x Sin 37⁰ = 1.06 N
% Uncertainty - Cos 37⁰ = (0.025/ Cos 37⁰) x 100 = 3.1 %
Uncertainty in F1x = 1.41 N x 3.1% = ± 0.044 N
F1x = -1.41 ± 0.044 N
% Uncertainty in Sin 37⁰ = (0.025/ Sin 37⁰) x 100 = 4.2 %
Uncertainty in F1y = 1.06 N x 4.2 % = ± 0.044 N
F1y = 1.06 ± 0.044 N
Uncertainty in Resultant = √((ΔFx)^2 + (ΔFy)^2)
Where ΔFx = Uncertainty in Fx = 0.044 N x 2 = 0.088 N
ΔFy = Uncertainty in Fy = 0.044 N x 2 = 0.088 N
Resultant Uncertainty = √((0.088 N)^2 + (0.088 N)^2) = 0.0624 N
F1 = 1.16 ± 0.876 N
Static Equilibrium of Forces Lab Report. (2016, Mar 11). Retrieved from https://studymoose.com/document/polygon-of-forces
👋 Hi! I’m your smart assistant Amy!
Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.
get help with your assignment
