A uniform metre rule is freely pivoted at the 11cm mark and it of mass 80g is hung from 2cm mark ?
a) draw a clear force diagram of the arrangement?
b) calculate the upright of the rule?
c) calculate the upload force extelide by the pivote
a) To draw a force diagram of the arrangement, you should consider the following forces:
1. Weight of the meter rule (acting downwards) - This can be represented by a downward arrow at the center of gravity of the meter rule.
2. Tension force (acting upwards) - This force arises because the 80g mass is hung from the 2cm mark. It can be represented by an upward arrow at the 2cm mark.
3. Pivot force (acting downwards) - This force arises due to the pivot supporting the meter rule. It can be represented by a downward arrow at the 11cm mark.
b) To calculate the weight (upright) of the meter rule, you need to determine the mass of the meter rule and then convert it to weight. Given that the mass of the meter rule is 80g, the weight can be calculated using the formula: Weight = mass x gravity, where gravity is approximately 9.8 m/s^2.
Weight = 0.08 kg x 9.8 m/s^2 = 0.784 N
Therefore, the upright force of the meter rule is approximately 0.784 Newtons.
c) To calculate the upload force exerted by the pivot, you need to consider the equilibrium condition of the meter rule. As the meter rule is freely pivoted, the sum of the anti-clockwise moments must be equal to the sum of the clockwise moments.
The anti-clockwise moments come from the tension force (0.08 kg x 0.02 m) and the upright force (0.784 N x 0.11 m). These moments result in a clockwise rotation. Since the meter rule is in equilibrium, the sum of the clockwise moments must be zero.
Therefore, the upload force exerted by the pivot can be calculated as follows:
Clockwise moments = 0
Upload force x distance from pivot = (0.08 kg x 0.02 m) + (0.784 N x 0.11 m)
Upload force x 11 cm = 0.0016 kg m^2/s^2 + 0.08624 kg m^2/s^2
Upload force x 11 cm = 0.08784 kg m^2/s^2
Upload force = 0.08784 kg m^2/s^2 / 0.11 m
Upload force = 0.8 N
Therefore, the upload force exerted by the pivot is approximately 0.8 Newtons.