Final FV1 Kress

Lab FV1 - Forces and vectors Equipment  Metal board with magnetic accessories: o Two pulleys o Force wheel with central floating disk  Set of hangers and masses Important (READ!!!)  To lift and move the magnetic components on the board, always handle the component by the magnetic base.  Magnetic elements that are not in use should be kept stuck to the back of the board Lab FV1 - Page 1

The apparatus The apparatus illustrated below will be used to study the relation between forces in an equilibrium condition. Find the force wheel with the floating disk in the center. Three strings should be already threaded through the center hole. If one or more are missing, please ask your instructor for assistance. Each of the strings should be used to support a hanger with masses as shown. The pulleys have very low friction, so the tension in the strings will be equal to the weight of the hanger + masses. Do NOT use the same masses left and right ( i.e ., do not use a symmetrical setting: it’s a boring one!). Start by placing mass on the center hanger, then add masses to the other hangers and adjust the angles of the strings by moving the pulleys until the disk at the center of the force wheel is more or less “floating”. Finally, lift the force wheel gently off the board and move it on the board as needed until the floating disk is perfectly centered with the force wheel. In this position, the disk is, indeed, floating: it is not in contact with the wheel. If we neglect the weight of the disk (which is indeed very small), the only forces acting on it are the three tensions in the strings. Lab FV1 - Page 2

Use the measured value of T 1 and θ 1 to calculate the numerical value of the components of . T 1x = T 1 *cos(20 o ) = 0.5978N * cos(20 o ) = 0.5617 N = T x T 1y = T 1 *sin(20 o ) = 0.5978N * sin(20 o ) = 0.2045 N = T y Uncertainty However, no measurement is perfect! In this lab we will not use standard but maximum errors, in order to simplify the calculations. The masses we are using have an uncertainty of  2% around their nominal value. A reasonable estimation of the uncertainty in the angles could be set to  1° (we can read the angle with a precision of about 0.5°, but we are increasing it to account for the friction in the system, which results in a certain variation in the position of the strings in equilibrium). Rewrite the values of your experimental values for T 1 and θ 1 , including uncertainty: M 1 = 0.061  0.0012  min= 0.0598 x 9.8 = 0.5860 N, max = 0.0622 x 9.8 = 0.6096 N T 1 = 0.5978  0.0118 θ 1 = 20 o  1 o Uncertainty is propagated through any calculation. Given the ranges of variation of T 1 and θ 1 , find the minimum and maximum values for T 1 y , the vertical component of the tension. Then, write the results as a central value ± error. T 1 = 0.5978  0.0118  [0.5860, 0.6096] θ 1 = 20 o  1 o  [19, 21] Max = sin(21 o )*0.6096 = 0.2185 N Min = sin(19 o )*0.5860 = 0.1908 N T 1y = 0.2047  0.0139 N Do the same thing for T 1 x , the horizontal component of the tension. Be careful, the value of the cosine increases as the angle decreases! T 1 = 0.5978  0.0118  [0.5860, 0.6096] θ 1 = 20 o  1 o  [19, 21] Max = cos(19 o )*0.6096 = 0.5764 N Min = cos(21 o )*0.5860 = 0.5471 N T 1x = 0.5618  0.0147 N Lab FV1 - Page 4

Experimental setup and background Let us go back to the whole system. The figure below shows the free-body diagram for the floating disk. (The mass of the disk is negligible.) The disk is in equilibrium. Therefore, according to Newton’s laws: [Equation 1] The magnitudes of these forces are equal to the hanging weights: Let us now rewrite vector equation [1] in terms of components, using the coordinate system in the figure. [Equation 2] Lab FV1 - Page 5

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Conclusions Does your data verify Newton’s second law, i.e. , equation 2? Your reasoning must include uncertainty. Yes! Our results verify the law because our F net values contain 0 with the uncertainty range! Lab FV1 - Page 8