the center of gravity of a 100-g stick is located at 50 cm mark and the stick is supported at 60cm. mark, Where must an 80g object be hung in order to have equilibrium?
summing moments about the center:
80x-10*100=0
x=1000/80=100/8 cm from the center on the support side.
thanks
12.5
To determine the position where the 80g object must be hung to achieve equilibrium, we need to consider the principle of torque and balance. Torque is the rotational force applied to an object.
In this scenario, the stick is being supported at the 60cm mark, and the center of gravity of the stick is at the 50cm mark. We know that torque depends on the distance from the point where the force is applied to the center of rotation.
First, let's calculate the torque exerted by the stick. The torque of an object can be calculated by multiplying the force applied to it by the distance from the center of rotation:
Torque (T) = Force (F) × Distance (d)
Since the stick is balanced, the torque generated by the stick must be equal and opposite when compared to the torque generated by the 80g object.
The torque exerted by the stick can be calculated as follows:
Torque of stick (T_stick) = Force of stick (F_stick) × Distance from support (d_stick)
Given that the weight of the stick is 100g, which is equivalent to 0.1kg, and the distance from the support is 10cm (60cm - 50cm), the torque exerted by the stick is:
T_stick = 0.1kg × 10cm
The torque exerted by the 80g object can be calculated as follows:
Torque of object (T_object) = Force of object (F_object) × Distance from support (d_object)
Given that the weight of the object is 80g, which is equivalent to 0.08kg, and we want to find the distance from the support, we can represent it as x cm. Therefore, the torque exerted by the object is:
T_object = 0.08kg × x cm
Since we want to achieve equilibrium, the torque exerted by the stick and the torque exerted by the object must be equal:
T_stick = T_object
Substituting the known values, we have:
0.1kg × 10cm = 0.08kg × x cm
Simplifying the equation, we can solve for x:
1 kg × 10cm = 0.8 kg × x cm
10cm = 0.8x cm
x cm = 10cm / 0.8
x cm = 12.5 cm
Therefore, the 80g object must be hung at the 12.5cm mark in order to achieve equilibrium.