gonzalezeduardo_4622624_288548587_Experiment 14 Torques, Equilibrium, and Center of Gravity-1

PHY2048L EXPERIMENT 14: Torques, Equilibrium, and Center of Gravity 94.0% Student name 1 Eduardo Gonzalez 2 Taylor Roca 3 Nadia Ramjohn 4 5 6

Experiment 14: Torques, Equilibrium, and Center of Gravity Page 2 of 7 A. Apparatus with Point of Support at Center of Gravity Measured/based on procedure Excel calculated 95.2 72.74 9.8 24.25 49.9 DATA TABLE 1 100% Note: Convert units! Diagram* Case 1 124.25 15 34.9 0.425 Paste in folder 224.25 69.5 19.6 0.431 Percent diff. 1% Case 2(a) 124.25 30 19.9 0.434 Paste in folder 224.25 70 20.1 0.442 74.25 23.6 26.3 Percent diff. 2% Case 2(b) 124.25 20 29.9 0.364 Paste in folder 224.25 60 10.1 0.365 74.25 69.5 19.6 Percent diff. 0% Case 3 1 48.9 67.590 Paste in folder 224.25 64.6 14.7 67.412 Percent error 0% Mass specified in procedure + m c Mass of meter stick ( g ) Total mass of 3 clamps ( g ) = g (m/s 2 ) Average mass of one clamp, m c ( g ) = Balancing position (center of gravity) of meter stick, x 0 ( cm ) = Values (add m C to masses if clamps used) units = grams Moment (lever) arms units = cm Results units = N-m m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW m 3 x 3 r 3 m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW m 3 x 3 r 3 x 1 (known) r 1 m 1 (measured) m 2 (known) x 2 (from expt) r 2 m 1 (calculated) * Draw a diagram to illustrate each case, using the Case 1 diagram as an example.

Experiment 14: Torques, Equilibrium, and Center of Gravity Page 3 of 7 B. Apparatus Supported at Different Pivot Points Measured/based on procedure Excel calculated 95.2 72.74 9.8 24.2466667 22.3 0.952 96% DATA TABLE 2 Note: Convert units! Diagram* 90% Case 5(a) 124.25 1 21.3 0.259 95.2 51 28.7 0.268 26 Torque difference 3% 85% Case 5(b) 124.25 1 21.3 0.259 73.97 48 25.7 0.268 21.23 61.5 39.2 Torque difference 3% 36.8 0.0000 100% Case 6(a) 100 1 29.5 0.289 100 60 29.5 0.289 30.5 Torque difference 0% 100% Case 6(b) 100 1 34.5 0.338 100 70 34.5 0.338 35.5 Torque difference 0% 100% Case 6(c) 100 1 39.5 0.387 100 80 39.5 0.387 40.5 Torque difference 0% 100% Case 6(d) 100 1 0.436 46.7 100 90 0.436 45.5 Percent diff. 2.60% Mass specified in procedure + m c Mass of meter stick ( g ) Total mass of 3 clamps ( g ) = g (m/s 2 ) Average mass of one clamp, m c ( g ) = Balancing position (center of gravity) of meter stick, x 0 ( cm ) = Linear mass density of meter stick, m = mass/Length ( g /cm) = Values (add m C to masses if clamps used) units = grams Moment (lever) arms units = cm Results units = N-m m 1 x 1 r 1 t CC m 2 = mass of the entire meter stick. m 2 x 2 r 2 t CW x' 0 m 1 x 1 r 1 t CC m 2 = mass of the meter stick to right of fulcrum m 2 x 2 r 2 t CW m 3 = mass of the meter stick to left of fulcrum m 3 x 3 r 3 x' 0 D t 5b - D t 5a m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW x' 0 m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW x' 0 m 1 x 1 r 1 t CC m 2 x 2 r 2 t CW x' 0 m 1 x 1 t CC x' 0 (measured) m 2 x 2 t CW x' 0 (predicted) * Draw a diagram to illustrate each case, using the Case 1 diagram as an example.

Experiment 14: Torques, Equilibrium, and Center of Gravity Page 4 of 7 85% QUESTIONS 100% 1. 100% 2. Why are clockwise and counterclockwise referred to as "senses," rather than directions? 50% 3. 50% 4. 100% 5. 220 5 35 120 90 50 40 Was the center of gravity of the meter stick exactly at the 50-cm mark? If not, why? Explain how the condition is satisfied for the meter stick in part A of the experiment. The meter stick was not balanced at 50 cm mark, indeed it was at the 49.9 cm mark. It is not necessary for a meter stick to balance at the 50 cm mark, which might be due to non uniformity of mass. To satisfy ΣF=0 , The fulcrum was placed on the center of gravity so the meterstick is balanced with a net force being 0. It means the fulcrum was moved to the point of balance in order to make the net force zero. Also, it is satisfied because the center of gravity, where the resultant weight of the body acts perpendicularly downward is counterbalanced by a number of forces acting upwards. Clockwise and counter-clockwise are considered as a sense rather than a direction because it is only a viewpoint where its movemement of direction is always changing. It is assumed as a tangential velocity of different points on the same body that is spinning in all directions and does not describe the motion of the object. You cannot point to a direction and say, it is turning this way, so you have to sue a sense as opposed to direction. Suppose the situation like Case 2(a) in the experiment, m 1 = 200 g were at the 20-cm position and m 2 = 100 g at the 65-cm position. Would there be a problem in experimentally balancing the system with m 3 = 50 g? Explain. If so, how might the problem be resolved? In order to balance the meter stick about its centre of gravity in the given scenario, the concept of equilibrium of torques must be used.According to the equilibrium of torque, the net torque on it must be zero. This is possible only when the total clockwise torques is equal to total anti-clockwise torques.I n the given scenario, the mass m1=200g is placed to the left of the 50 cm mark and mass2= 100g is placed to the right of the 50 cm mark. The distance of mass from the centre of mass is, x1= 50cm -20cm; x1=30 cm The distance of mass from the centre of mass is, x2=65cm-50 cm; x2=15 cm Considering that the third mass (m3=50) is placed at an arbitrary position X3 from the centre of gravity and applying the equilibrium of torque about the center of gravity and consider the clockwise toque as negative and counter-clockwise torque as positive. Describe the effects of taking the mass of the meter stick into account when the balancing position is not near the 50-cm position. If the equilibrium position is not directly in the center,we do not have an evenly distributed mass, we cannot simply assume gravity "acts" at the center of the stick. We have to provide additional calculations in order to determine the point where we can establish gravity "acts." For instance, if we have the equilibrium point at 52 cm, this means we have more "dense" mass from 52 cm to 100 cm, and the system will move in this direction.(assuming our axis of rotation is still in the center of the stick Without this shift in equilibrium point from the axis of rotation, we would have no "net" torque, because if there was no shift, the weight would "act" on the axis of rotation itself. Therefore, there would be no angular displacement or velocity without this above situation occuring in some form. (Optional: Extra credit up to 10 points) A uniform meter stick is in static rotational equilibrium when a mass of 220 g is suspended from the 5.0-cm mark, a mass of 120 g is suspended from the 90-cm mark, and the support stand is placed at the 40-cm mark. What is the mass of the meter stick? m1*g*r1-m3*g*r3-m2*g*r2=0 m1*g*r1-m2*g*r2=m3*g*r3 (m1*r1-m2*r2)(r3)=m3 m3=(((0.220 kg)*(0.35 m))-((0.120 kg)(0.5 m)))/(0.1 m) m3=0.170 kg m3=170 g m 1 x 1 r 1 m 2 x 2 r 2 x 0  F i = 0

m 1 10 20 30 0 r 1 x 1