I know that for the second exercise the same rule basically applies, and if I solve knowing the things I just wrote I will obtain:
The lever_arm is not l/2, but it's (l/2-d) or, in other terms, the lever_arm is the distance between the object and the new balance point. So let's say I put the fulcrum at the original balance point of the stick: now d 1 is = d, whereas d 2 is (since the object is at the end of the stick) l/2. Forces are m 1*g and m 2*g, and so far so good. Why can't we use the same "torque formula" for both? In other words, if I use m 1*g*d=m 2*g*lever_arm I'll get the right result, but if I use m 1*g*d=m 2*g*d 2 (where d 2 is l/2) I get the wrong result.Īs far as I know, torque is generally defined by that Force*Distance equation, where "Distance" is the distance from the Force to the fulcrum, or rotation axis. The torques are opposed to one another since the system is in equilibrium, so solve the equation that way (m 2 = m 1*d/lever_arm) Calculate Torque by the object using: m 2*g*lever_arm Calculate Torque by the weight force of the stick using: m 1*g*d
#TORQUE EQUILIBRIUM AND CENTER OF GRAVITY LAB REPORT HOW TO#
I already know how to solve each problem, but I don't know why they are solved the way they are.īy playing around with random equations and googling similar problems, the answer to the first problem is the following: Stick of known length l, unknown mass m 1, balance point in the middle.īy placing an object of known mass m 2at a distance d 1 away from one end of the stick, the balance point moves towards the same end as the object by a distance of d 2. Theoretical (6 ) Case 2: Theoretical, Case 3: Theoretical For the Results and Conclusion sections, include a comparison of experimental and theoretical results.There is a stick of known length l and known mass m 1 with its balance point in the middle.īy placing an object of unknown mass m 2 at the far end of the stick, the balance point moves towards the same end as the object by a distance of d. m e) (cm) m(s) (cm) m(B) x (cm) Case I 15 NA NA Case 2 3 0 70 Case 3 10 90 NA NA (cm) x=, Laboratory 9: Torque, Equilibrium, and the Center of Gravity Find the location of the center of mass, ie, the point where the stick will balance Data and Data Analysis - center of mass of meter stick= m = mass of meter stick m = mss of hanger clamp Note: Form, my, and m, remember to include the mass of the hanger clamp, m. Place 100 g at the 10 cm position and 200 g at the 90-cm position. Repeat the activities of 2, except use the following masses, 100 g at the 30-cm position and 200 g at the 70-cm position Suspend m = 50 and find y, so this mass will balance out the meter stick. If you use the hanger clamps to suspend the masses, remember to include their masses in m 3. Suspend a second mass, 200 g on the opposite side of the stick as my, and place it at the distances needed to balance the meter stick. With the meter stick on the support stand at suspenda mas 100 g at the 15-em position on the stick.
Tighten the clamp and record the distance of the balancing point from the zero end of the meter stick 2. Place a knife-edge clamp on the meter stick at the 50-cm lime and place the meter stick on the support and Ad the meter stick through the camp at the meet stick is balanced on the stand. Transcribed image text: Laboratory #9 Report Sheet: Torque, Equilibrium and Center of Gravity Objective The purpose of this laboratory is the Tid bodies mechanical equilibrium toque and how it apo Suggested Procedure 1.
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