Our counterweight was composed of the entire aluminum U channel with the 6” x 6” x 0.025” aluminum sheet folded up and bolted to the side. In total, it weighed approximately 5.2 oz. The counterweight was mounted at a 30 degree angle onto the delrin strip, at a point 5.5 inches from the motor. We neglected the mass of the delrin strip because the pivot point about the servo was located directly in the middle of the delrin strip. This meant that the weight of the delrin strip could not produce a moment about the servo, and thus was not necessary in our calculations of torque. We assumed the center of mass of the counterweight to be halfway along its 6 inch length, and that the force of the block to be lifted acts 5.5 inches away from the motor.
By taking the moment about C, we can calculate the torque that must be exerted by the motor under 2 conditions: When the lever is only holding up the counterweight and when the lever is lifting the block.
Case 1: Only holding up the counterweight
∑M|C = (5.2 oz)(5.5 in + 3cos(30) in) + Mmotor = 0
Mmotor = -42.11 oz-in
The negative sign signifies that the torque supplied by the motor must act in the opposite direction than shown to support the counterweight. Since the maximum torque that the motor can output is 57 oz-in, the motor is able to support the counterweight initially.
Case 2: Lifting the block
∑M|C = (5.2 oz)(5.5 in + 3cos(30) in) + Mmotor – (16 oz)(5.5 in) = 0
Mmotor = 45.89 oz-in
The maximum torque that the motor can output is 57 oz-in, which means the motor is able to lift the block.
Theoretical Calculations
Now that we determined that our mechanism is stable initially, we want to determine how high it can lift the weight. To calculate the max lift height, we determined the angle at which the torque supplied by the motor and the counterweight would equal the torque produced by the weight. With this angle, we can find the displacement of the weight using geometry.
∑M|C = (5.2 oz)(5.5 in + 3cos(30) in)(cos(Ï´)) + (57 oz-in) – (16 oz)(5.5 in) = 0
Ï´ = 42.6 o
h = (5.5in)cos(42.6)
h = 3.72in
​
We predict that our crane can lift the weight 3.72 inches. This prediction is very close to our observed result of 3.25 inches. The discrepancy can be accounted for in the bending of the crane arm. Using SolidWorks FEA, we determined that the 16 oz load will displace the farthest end by about 1cm or 0.4 inches. With this consideration included, our prediction is nearly identical to the observed result.