The working for a body of mass 50.0cm is suspended at the 20.0cm mark of a uniform metre rule. The metre rule is adjusted on a pivot until it settles horizontally at the 40.0cm mark. Determine the mass of the metre rule.
Yes
up force at 40 cm = (m+50)g
down force at 20 cm = 50 g
down force at 50 cm = m g, the weight of the stick
moments about the 40 cm mark are 0
so
50 (40-20) = m (50 -40)
1000 = 10 m
m = 100
To determine the mass of the meter rule, we need to consider the principles of rotational equilibrium.
Let's break down the problem step-by-step:
1. We have a body with a mass of 50.0 g suspended at the 20.0 cm mark of the meter rule. This implies that the weight of the body (W) acts downward at the center of mass of the body, which is at the 20.0 cm mark.
2. The meter rule is adjusted on a pivot until it settles horizontally at the 40.0 cm mark. When the meter rule is horizontal, it means that the total clockwise torque (moment) is balanced by the total counterclockwise torque (moment). In other words, the system is in rotational equilibrium.
Now, let's find the mass of the meter rule using the principle of torque equilibrium:
1. The torque (τ) of an object is given by τ = r * F * sin(θ), where r is the perpendicular distance from the pivot to the line of action of the force, F is the force, and θ is the angle between the line of action of the force and the lever arm (perpendicular distance).
2. The meter rule is a uniform object, meaning that its mass is distributed evenly along its length. Therefore, the center of mass of the meter rule is located at its midpoint, which is the 50.0 cm mark.
3. The weight of the meter rule (W_rule) can be considered as acting downwards at the center of mass, which is at the 50.0 cm mark. To balance the system in rotational equilibrium, there must be an equal and opposite torque acting around the pivot.
4. The torque due to the weight of the meter rule (τ_rule) is given by τ_rule = r * W_rule * sin(θ), where r is the distance of the center of mass from the pivot (which in this case is 40.0 cm).
5. Since the system is in rotational equilibrium, the sum of clockwise torques must be equal to the sum of counterclockwise torques. Therefore, τ_rule = τ_body.
6. Now we can solve for the mass of the meter rule. We know that the weight (W) of an object is given by W = m * g, where m is the mass and g is the acceleration due to gravity.
7. Using the torque equation and equating the torques: r * W_rule * sin(θ) = r * W * sin(θ)
8. Since we are looking for the mass of the meter rule, we need to express the weight of the meter rule in terms of its mass. We can use the equation W = m * g and rewrite it as W_rule = m_rule * g, where m_rule is the mass of the meter rule.
9. Substituting the expression for the weight of the meter rule into the torque equation: r * (m_rule * g) * sin(θ) = r * (m * g) * sin(θ)
10. Now we can cancel out the common factors (r and sin(θ)) from both sides of the equation: m_rule * g = m * g
11. Finally, dividing both sides of the equation by g (acceleration due to gravity): m_rule = m
So the mass of the meter rule is equal to the mass of the suspended body, which is 50.0 g.
Assuming you meant 50.0g (not cm), then consider the mass of each part of the bar as a point mass at the midpoint of its length. Then if the bar's mass is x, then
(.4x+50)(20) = (.6x)(30)