Coefficient of Friction (µ) is the ratio between the force necessary to move one surface horizontally over another and the pressure between the two surfaces. Normal Force (N) is a support force exerted upon an object that is in contact with another stable object. It is always perpendicular to the surface of the object. THEORY Treating both masses together as one system, the free-body diagram includes two forcers: the force pf gravity putting on mass m and the kinetic friction acting on mass m. According to Newton’s 2nd law, the vector sum of these forces equals the mass of the system times the acceleration of the block. ∑ F = mg - fK = (M + m) a (1) By deriving the equation above, we can get the
Displace the mass by 0.02, 0.04, 0.06, 0.08, 0.10 meters 6. Use Labpro to measure the velocity of the spring while in motion 7. Record the maximum positive velocity in a data table Data Collection: Part 1: |Cord
Before the impact of the fluid onto the plate, the fluid is in line with the x-axis, as shows by the velocity vector labeled V1. After the water impacts with the plate the velocity vector V2 is the parallel to the x-axis. The equation to fiqure out the normal force on the plate is: Fx=ρQV1 Where ρ Is density, Q = flow rate and V1 is the x-component of velocity Force measured (1.4) this is a diagram of the weight arm and pivot used in this experiment. Using this diagram two equations can be found and used. When the valve is closed, no water flows and so the weight is positioned at distance L from the pivot.
Strengths Lab report Calculating deflection of a statically indeterminate Beam under loading Summary The purpose of this experiment was to investigate the accuracy of theoretical calculation for deflection in beams compared to that of experimental procedure. Our findings showed us that theoretical calculations are an accurate method of approximation and also that deflection directly varies with transverse force applied. Therefore the use of theoretical calculations for engineering design is an accurate and useful and time saving method. We also used load cells to calculate reaction forces and moments. For the calculation of the reaction moment at the support mounted on an arm, the value for the reaction force is multiplied by the distance at which the force is acting.
One newton is defined as the resultant force, which causes a mass of 1kg to accelerate at 1ms-2. Acceleration=Unbalanced force/mass (or F= ma) therefore newton’s second law tells us that force causes acceleration. Drag force acts in opposition to the direction of movement (Mark W. Denny,1993). Drag force depends on density (ρ) of the medium in which the object is moving, the cross-sectional area in the direction of movement, the drag coefficient (depends on shape and surface properties) and the velocity. It is in Newton’s second law that drag force decelerates and performs negative work (University of Glasgow, 2015).
Then, a dynamic absorber comprising a thin and flexible beam with two masses which can slide along the flexible beam was mounted onto the top of the concentrated mass. In order to optimize the use of the absorber and reduce the vibration to minimum, the specific position of the masses of the dynamic absorber was determined where minimum vibrating amplitude was detected. Lastly, the frequency response function of the structure with dynamic absorber was recorded and compared with that of the original one. 2. Introduction Machines usually operate at a fixed speed and exhibit vibration, so the machines will generally be excited at a fixed frequency which is sometimes the operating frequencies or other fixed frequencies which are proportional to the driving speed.
Newton's first law of motion states that objects at rest remain at rest unless an unbalanced force is applied. The second law of motion describes what happens if the resultant force is different from zero. If the acceleration is constant, the body is said to be moving with uniformly accelerated motion. The purpose of this experiment is to measure the acceleration of a given mass produced by a given force and compare it with that calculated from Newton's second law of motion.
Figure 1 2. The solid shaft has a diameter of 0.75 in. If it is subjected to the torques shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and F allow free rotation of the shaft. Figure 2 3.
0 = Ma + kx , rearrange a = -(k/m)x The acceleration is directly proportional to displacement, and it is directed towards the rest position. If the x was plotted against time,a sinusoidal graph would result. The constants proportionally is k/m and this must be the angular frequency squared so wn=km. The frequency of oscillation is f = w/2π =1/2πkm The natural frequency are often denoted as wn and fn.This equation is true for all elastic oscillation \ EXPERIMENTAL-Related formula Equation of motion (EOM), Mo=Jo φ=-Fca Fc=cx=cφa Where, Mo is moment about pivot point o of the beam Fc is spring force,result of the deflection x and spring constant c For small angles, the deflection can be formed from the torsion φ and level arm a Jo=mL33 Where, Jo is mass moment of inertia of the beam about pivot EOM in the form of homogenous differential equation, φ+ 3ca2mL2
The reading was recorded and the procedure was repeated with mass (m) 75 g and 100 g. While for the third case, the mass and speed are kept constant at N = 150 rpm and m= 50 g. The radii value was recorded first at 20 mm and was repeated for radii 45 mm, 70 mm and 95 mm respectively. 4. Theory Centrifugal force formula comes from the Newton’s 2nd Law of motion which states that a body moving in a circular path with constant speed must be acted upon by an unbalanced force which is always directed towards the centre. This necessary unbalanced force is called the centripetal force. We can simply put it as F = ma.