The objective of the experiment is to observe how the weight of the beam on which the forces act behaves like a force concentrated at the center of gravity of the beam and to determine the conditions of equilibrium for several parallel forces.

Before the experiment proper, Torque was discussed extensively. The experiment is divided into two parts. In the first experiment, the meter stick was weighed and the group was tasked to find the center of gravity of the uneven meter stick and supporting the meter stick other than its center of gravity, bring it into equilibrium using the weight of a single mass.

In the second part of the experiment, the meter stick was supported on the 30 cm mark and weights were applied (200g at 10 cm mark, 100g at 20cm mark, 20g at 90 cm mark). The group was tasked to locate the position at which the 50 g must be hanged to produce rotational equilibrium. The group was able to compute for the torque arm and small percentage error occurred.

GUIDE QUESTIONS AND PROBLEMS

1. Define the following:

a. The moment of a force – it is the tendency to cause rotation about a point or an axis. It is also called torque (τ).

b. The center of gravity – it is the point where all the weight of an object can be considered to be concentrated. 2. Differentiate transitional and rotational equilibrium

Rotational equilibrium happens when the sum of all torques acting on an object is zero while transitional equilibrium happens when an object has no or zero net force acting on it.

3. A uniform rod 1m long weighs 150N and is supported on some fulcrum. Weights of 40N and 50N are suspended from the two ends of the rod. Find the position of the fulcrum if the system is in equilibrium.

...Torques equilibrium, and center of gravity
Introduction
Torque is a quantitative measure of the tendency of a force to cause or change the rotational motion of a rigid body. A torque is the result of force acting at a distance from an axis of rotation. An essential thing to keep in mind is that the magnitude of the torque is equal to the product of the forces perpendicular distance and magnitude.
Theory
The magnitude of the torque (t) is found from the product of the force F and the perpendicular distance from the axis of rotation to the forces line of action. When there is no net torque acting on a stationary rigid body, the body will then be in static rotational equilibrium and there is no rotational motion.
Materials
* meterstick
* support stand
* laboratory balance
* string
* knife-edge clamp/four knife-edge clamps
* four hooks
* weights
* unknown mass with hooks
Procedure
First the apparatus was to be set up which consisted of a supporting stand, meterstick, and knife edge clamp. The mass of the clamp and meterstick were measured on a balance beam. The clamps and weight hangers were attached to the meterstick and the instructed weights were added to each hanger and the clamps were situated on the accurate spot. The...

...Torques
Introduction During this lab you will become more familiar with the concepts of torque. The purpose of this lab is to determine if the rotational equilibrium condition, Στ = 0, holds experimentally. Equipment Meterstick (1) - no metal ends Fulcrum (1) Clamps (4) Weight Hanger (1) Mass Set (1) Digital Scale (1)
Theory For a body to be in static equilibrium, two conditions have to be met:
ΣF = 0 and Στ = 0
where F is force and τ is torque. (The torque is the force times the lever arm, r) The first condition, ΣF = 0, is concerned with translational equilibrium and ensures that the object is at rest or is moving at a uniform linear velocity. The second condition, Στ = 0, is concerned with rotational equilibrium and ensures that the object is not rotating or is rotating at a uniform angular velocity. Special Instructions Goals - One of the primary goals of this lab is accurate measurement with attention to detail and the use of the correct number of significant figures. I know that you can finish this lab quickly – but don’t. Squeeze as much precision as you can out of this set of equipment. Think up ways of avoiding errors and inaccuracies rather than explaining them away in the lab report. Units - In this lab you will find it convenient to use grams as the mass unit and centimeters as the distance measure. The nature of the torque balance equations allows us to...

...FLOW METER
There are well over 20 different types of flow meters, even if we lump the various positive-displacement flow meters together as one type. Unless the process engineer knows the pros and cons of each type, it can be a daunting task to properly select one. Here are just some of the factors to consider before selecting a flow meter:
• Its size and measuring range of the flow meter
• Chemical compatibility
• Process accuracy requirements• Pressure requirements
• Acceptable pressure drop
• Cleaning requirements (i.e., do you need, and does the unit offer,
Clean-in-place capabilities?)
• Desired measurement units (such as volume, velocity or mass)
• Uni-directional or bi-directional measurement
• Fluid viscosity limitations
• Necessary approvals for use in hazardous areas, sanitary
applications and so on (examples include Factory Mutual, Canadian Standards Assn., 3-A Standards and Accepted Practices, and Underwriters’ Laboratory approvals)
• Custody-transfer approvals
• Data-output requirements (i.e., 4–20 mA, relay, digital or simple display)
• Calibration and re-calibration requirements
• Maintenance issues
• Operating costs
• Connection styles (flanged, wafer, threaded, weld-on and so on)
MAGNETIC FLOW METER
A Magnetic flow meter is a volumetric flow meter which does not have any moving parts and is...

...Analysis
Torque is the ability of force to change the rotational motion of a particle. It is also called the moment of force. It is always specified with regard to the axis of rotation. On the experiment the axis of rotation serves as the model balance. This means that as much as torque is directly proportional with the force applied on a particle, it is also dependent on the perpendicular distance of the applied force to the axis of rotation. On the first activity we need to determine the weight of the pans. At first we had a difficulty or rather error on the activity because we put weights on both of the pans which causes the equilibrium to be invalid. On the third activity we need to use the second hole in the beam as the axis of rotation so that the center of gravity of the beam does not pass through the new axis of rotation.
Conclusion:
The experiment was done with the purpose to analyze systems in equilibrium using the second condition of equilibrium and to distinguish some of its use and significance. In the experiment we analyzed systems and how it is applied. After performing the experiment, I can therefore conclude that the torque is affected by the forces acting on the system and their radial distance from the axis of rotation, the shorter the lever arm the greater the force, the longer the lever arm the lesser the force.
Related Discussion
According to this condition, an object that is...

...pivoted about an axis through point O at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces F1 = 18 N,
F2= 26N, and F3= 14N.
12. The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8 s, starting from rest?
13. A uniform 255 N rod that is 2m long carries a 225 N weight at its right end and an unknown weight W toward the left end. When W is placed 50 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75 cm form the right end.
a. Find W.
b. If W is now moved 25 cm to the right, how far and in what direction must the fulcrum be moved to restore the balance?
14. A 1.5 kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during that time? (c) Calculate the work done by the torque. (d) What is the grinding wheel’s kinetic energy when it is rotating at 1200 rev/min?
15. A 50 kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N. The coefficient of kinetic friction between the blade and the stone is 0.60 and there is a constant friction torque of 6.50 N-m between the...

...UNIVERSITY
DEPARTMENT OF PHYSICS
LAB REPORT
FOR
PCS 211 SECTION **
EXPERIMENT: STATIC EQUILIBRIUM - FORCES AND TORQUES
EXPERIMENTERS: ***** *********
***** *********
AUTHORS OF THIS REPORT ***
***
EXPERIMENT PERFORMED ON: ***
REPORT SUBMITTED ON: ***
INSTRUCTOR: ***
PRE-LAB QUESTIONS:
1) What is meant by static equilibrium?
The meaning of static equilibrium can be explored by first examining the definition of equilibrium. Equilibrium means that an object is at rest or that the objects center of mass moves at constant velocity relative to the observer.
Static equilibrium is the case where an object is at rest. An object is at rest or at static equilibrium when the net force applied on it is zero. When an object is at rest about an axis it may have a tendency to rotate depending on the distribution of mass or if a force is applied. This tendency to rotate is called torque must also be equal to zero for the conditions of static equilibrium to be satisfied.
2) How do you define torque (or moment of a force)? Give the most general definition, illustrate with a diagram.
Force
Torque is the tendency of a force to rotate an object about an axis. Torque is defined as the product of the force applied perpendicularly to the surface of the affected object and the distance of that force from a point of pivot –
Torque = Force *...

...diameter of a square threaded spindle of a screw jack is 40 mm. The screw pitch is
10 mm. If the coefficient of friction between the screw and the nut is 0.15, neglecting friction
between the nut and collar, determine
(i) Force required to be applied at the screw to raise a load of 2000 N
(ii) The efficiency of screw jack
(iii) Force required to be applied at pitch radius to lower the same load of 2000 N and
(iv) Efficiency while lowering the load.
J
3
An open belt drive connects two pulleys 120 cm and 40 cm diameter on parallel shafts 4 m
apart. The maximum tension in the belt is 1855 N. The coefficient of friction is 0.3. The
driver pulley of diameter 120 cm runs at 200 rpm. Calculate:
(i) The power transmitted.
(ii) Torque on each of the two shafts.
4
A steel ball of diameter 150 mm rests centrally over a concrete cube of size 150 mm.
Determine the center of gravity of the system, taking weight of concrete = 25,000 N/m2 and
that of steel 80,000 N/m 2.
Contd. in Page 2
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R5
Code: R5 100305
5
(a)
(b)
6
(a)
(b)
7
(a)
(b)
8
Show that the moment of inertia of a thin circular ring of mass ‘M’ and mean radius ‘R’
with respect to its geometric axis is MR2.
Find the mass moment of inertia of a right circular cone of base radius ‘R’ and mass ‘M’
about the axis of the cone.
A train is uniformly accelerated and passes successive...

...WHAT IS TORQUE?
Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'. The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by 'r'. Note that this distance, 'r', is also a vector, and points from the axis of rotation to the point where the force acts. (Refer to Figure 1 for a pictoral representation of these definitions.) |
Figure 1 Definitions |
Torque is defined as
= r x F = r F sin().
In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F.
Using the right hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector.
Imagine pushing a door to open it. The force of your push (F) causes the door to rotate about its hinges (the pivot point, O). How hard you need to push depends on the distance you are from the hinges (r) (and several other things, but let's ignore them now). The closer you are to the hinges (i.e. the smaller r is), the harder it is to push. This is what happens when you try...

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