The objective of the experiment is to observe how the weight of the beam on which the forces act behaves like a force concentrated at the center of gravity of the beam and to determine the conditions of equilibrium for several parallel forces.
Before the experiment proper, Torque was discussed extensively. The experiment is divided into two parts. In the first experiment, the meter stick was weighed and the group was tasked to find the center of gravity of the uneven meter stick and supporting the meter stick other than its center of gravity, bring it into equilibrium using the weight of a single mass.
In the second part of the experiment, the meter stick was supported on the 30 cm mark and weights were applied (200g at 10 cm mark, 100g at 20cm mark, 20g at 90 cm mark). The group was tasked to locate the position at which the 50 g must be hanged to produce rotational equilibrium. The group was able to compute for the torque arm and small percentage error occurred.
GUIDE QUESTIONS AND PROBLEMS
1. Define the following:
a. The moment of a force – it is the tendency to cause rotation about a point or an axis. It is also called torque (τ).
b. The center of gravity – it is the point where all the weight of an object can be considered to be concentrated. 2. Differentiate transitional and rotational equilibrium
Rotational equilibrium happens when the sum of all torques acting on an object is zero while transitional equilibrium happens when an object has no or zero net force acting on it.
3. A uniform rod 1m long weighs 150N and is supported on some fulcrum. Weights of 40N and 50N are suspended from the two ends of the rod. Find the position of the fulcrum if the system is in equilibrium.
ΣT = T40N + T150N – T50N
Since ΣT = 0
0 = (40N)(x) + (150N)(x - 0.5) + (150N)(1 - x)
0 = 40x + 150x - 75 - 50 + 50x
0 = 240x – 125
240x = 125
X = 0.5208m or 52.08cm
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