a uniform meter stick is at rotational equilibrium when 220 g is suspended at 5 com, 120 g is suspended at 90, and the support stand is placed at the 40 cm mark what is the mass of the meter stick?
220*(40 - 5) - 120*(90 - 40) - m*(50 - 40) = 0 (the center of mass of the stick is at the 50 cm point).
m = [220*(40 - 5) - 120*(90 - 40)]/(50 - 40) = 170 g
To find the mass of the meter stick, we can use the concept of rotational equilibrium. In rotational equilibrium, the sum of the clockwise torques is equal to the sum of the counterclockwise torques.
Let's denote clockwise torques as negative and counterclockwise torques as positive.
Given:
Mass suspended at 5 cm = 220 g
Mass suspended at 90 cm = 120 g
Support stand at the 40 cm mark
First, let's calculate the torques due to the masses suspended:
Torque due to 220 g at 5 cm:
Torque1 = (220 g) * (9.8 m/s^2) * (0.05 m)
Torque1 = 107.8 N*m
Torque due to 120 g at 90 cm:
Torque2 = (120 g) * (9.8 m/s^2) * (0.9 m)
Torque2 = 105.84 N*m
Next, let's calculate the torque due to the meter stick itself:
Torque due to the meter stick at the 40 cm mark:
Torque3 = Mass of meter stick * (9.8 m/s^2) * (0.4 m)
Since the meter stick is at rotational equilibrium, the sum of the torques must be zero:
Total torque = Torque1 + Torque2 + Torque3
Since the total torque is zero, we can solve for the unknown mass of the meter stick:
0 = Torque1 + Torque2 + Torque3
0 = 107.8 N*m + 105.84 N*m + Torque3
Rearranging the equation:
Torque3 = -(107.8 N*m + 105.84 N*m)
Now, let's substitute the values and solve for the mass of the meter stick:
-(107.8 N*m + 105.84 N*m) = Mass of meter stick * (9.8 m/s^2) * (0.4 m)
Simplifying the equation:
-213.64 = Mass of meter stick * 3.92
Dividing both sides of the equation by 3.92:
Mass of meter stick = -213.64 / 3.92
Mass of meter stick ≈ -54.5 g
Note: The negative sign indicates that the meter stick would need a mass of 54.5 g at the 40 cm mark to achieve rotational equilibrium. This could imply that there is an error or some other factor affecting the equilibrium condition.
To find the mass of the meter stick, we can set up an equation based on the principle of rotational equilibrium. The principle states that the sum of the torques acting on a meter stick in rotational equilibrium is zero.
First, let's consider the 220 g weight suspended at the 5 cm mark. The torque exerted by this weight is given by the product of its weight and the distance from the pivot point (support stand in this case). Since the weight is suspended at 5 cm, the distance from the support stand is 40 cm - 5 cm = 35 cm.
The torque due to the 220 g weight can be calculated as:
Torque1 = (mass1) * (acceleration due to gravity) * (distance1)
Substituting the given values:
Torque1 = (0.220 kg) * (9.8 m/s^2) * (0.35 m)
Next, let's consider the 120 g weight suspended at the 90 cm mark. The distance from the support stand can be calculated as:
Distance2 = 40 cm - 90 cm = -50 cm (negative sign indicates direction relative to the support stand)
The torque due to the 120 g weight can be calculated as:
Torque2 = (mass2) * (acceleration due to gravity) * (distance2)
Substituting the given values:
Torque2 = (0.120 kg) * (9.8 m/s^2) * (-0.50 m)
According to the principle of rotational equilibrium, the sum of the torques should be zero. Therefore, we can write the equation as:
Torque1 + Torque2 = 0
Substituting the calculated values, the equation becomes:
(0.220 kg) * (9.8 m/s^2) * (0.35 m) + (0.120 kg) * (9.8 m/s^2) * (-0.50 m) = 0
Now, we can solve this equation to find the unknown mass of the meter stick.
0.220 kg * 9.8 m/s^2 * 0.35 m + 0.120 kg * 9.8 m/s^2 * -0.50 m = 0
0.220 kg * 3.43 N*m + 0.120 kg * -4.90 N*m = 0
0.7546 N*m - 0.588 N*m = 0
0.1666 N*m = 0.588 N*m
Therefore, the mass of the meter stick is determined by the torque equation. It is important to note that in this scenario, the other forces acting on the meter stick, such as the upward force from the support stand, do not contribute to the torque equation and can be ignored.