A rigid rod of mass
m3is pivoted at point A, and masses
m1 and m2 are hanging from it, as shown in the figure.
(a) What is the normal force acting on the pivot point? The ratio of the weights of
m1, m2, and m3 is 1:2:3. (Use the following as necessary: m1 and g for the acceleration due to gravity.)
N =
(b) What is the ratio of
L1 to L2,where these are the distances from the pivot point to
m1 and m2,respectively? (Use any variable or symbol stated above as necessary.)
L1/L2 =
(a) The normal force acting on the pivot point can be found by considering the torques acting on the rod. Since the rod is in rotational equilibrium, the sum of the torques must be zero. The torques in this case are due to the weights of m1, m2, and m3.
The torque due to m1 is τ1 = m1 * g * L1, where L1 is the distance from the pivot point to m1.
The torque due to m2 is τ2 = m2 * g * L2, where L2 is the distance from the pivot point to m2.
The torque due to m3 is τ3 = m3 * g * (L1 + L2), as the entire weight of m3 acts at the center of mass of the rod.
Since the rod is in rotational equilibrium, the sum of the torques is zero: τ1 + τ2 + τ3 = 0.
Using the ratio of the weights, m1 : m2 : m3 = 1 : 2 : 3, we can express m2 and m3 in terms of m1:
m2 = 2m1 and m3 = 3m1.
Substituting the expressions for m2 and m3 into the torque equation:
m1 * g * L1 + 2m1 * g * L2 + 3m1 * g * (L1 + L2) = 0.
Simplifying the equation:
m1 * g * L1 + 2m1 * g * L2 + 3m1 * g * L1 + 3m1 * g * L2 = 0.
5m1 * g * L1 + 5m1 * g * L2 = 0.
Dividing both sides by 5m1g:
L1 + L2 = 0.
L1 = -L2.
Since the lengths cannot be negative, we take L1 = L2 = L.
Therefore, the normal force acting on the pivot point is:
N = m1 * g + m2 * g + m3 * g = (m1 + m2 + m3) * g = (1 + 2 + 3)m1 * g = 6m1 * g.
So, N = 6m1 * g.
(b) The ratio of L1 to L2 is L1/L2 = 1/1 = 1.
To solve this problem, let's go through it step by step.
(a) To find the normal force acting on the pivot point, we need to consider the forces acting on the rod. Since the rod is in equilibrium, the sum of the torques and forces acting on it must be equal to zero.
Let's denote the weight of m1 as W1, the weight of m2 as W2, and the weight of m3 as W3. Since the ratio of the weights is 1:2:3, we can write:
W1 = m1 * g,
W2 = 2 * (m1 * g),
W3 = 3 * (m1 * g),
where g represents the acceleration due to gravity.
Now, let's consider the forces acting on the rod. We have the weight W1 acting downwards at a distance L1 from point A, the weight W2 also acting downwards at a distance L2 from point A, and the weight W3 acting downwards at a distance L3 from point A.
To find the normal force N, we can sum up the torques about point A. Since the rod is in equilibrium, the total torque must be zero:
W1 * L1 - W2 * L2 - W3 * L3 = 0.
Substituting the weights with their respective values:
(m1 * g) * L1 - (2 * m1 * g) * L2 - (3 * m1 * g) * L3 = 0.
Dividing through by m1 * g, we get:
L1 - 2L2 - 3L3 = 0.
Now, we need one more equation to solve for the normal force N. Since the rod is in equilibrium, the sum of the forces in the vertical direction must be zero:
N - W1 - W2 - W3 = 0.
Substituting the weights:
N - (m1 * g) - (2 * m1 * g) - (3 * m1 * g) = 0.
Simplifying:
N - 6(m1 * g) = 0.
Now, we have two equations:
L1 - 2L2 - 3L3 = 0, (Equation 1)
N - 6(m1 * g) = 0. (Equation 2)
From Equation 2, we can find the value of N:
N = 6(m1 * g).
Therefore, the normal force acting on the pivot point is N = 6(m1 * g).
(b) To find the ratio of L1 to L2, we can rearrange Equation 1:
L1 = 2L2 + 3L3.
Now, we can substitute this equation into Equation 2 to eliminate L1 and solve for L2:
N = 6(m1 * g) = 2L2 + 3L3.
Simplifying:
2L2 = 6(m1 * g) - 3L3,
L2 = 3(m1 * g) - (3/2)L3.
To find the ratio L1/L2, we can divide both sides of Equation 1 by L2:
L1/L2 = (2L2 + 3L3)/L2.
Simplifying:
L1/L2 = 2 + (3L3/L2).
Substituting the expression for L2:
L1/L2 = 2 + (3L3/(3(m1 * g) - (3/2)L3)).
Simplifying:
L1/L2 = 2 + (3L3/(3m1 * g - (3/2)L3)).
Therefore, the ratio of L1 to L2 is L1/L2 = 2 + (3L3/(3m1 * g - (3/2)L3)).
To find the normal force acting on the pivot point, we need to consider the forces acting on the system. We have two masses, m1 and m2, hanging from the rigid rod, and the rod has mass m3. The ratio of the weights of the masses is given as 1:2:3.
(a) We start by finding the weight of each mass. The weight of an object is given by the formula:
Weight = mass * acceleration due to gravity (g)
For m1, the weight is m1 * g.
For m2, the weight is m2 * g.
For m3, the weight is m3 * g.
Since the ratio of the weights is 1:2:3, we can write the weight of m1, m2, and m3 as follows:
Weight of m1 = (1/6) * total weight
Weight of m2 = (2/6) * total weight
Weight of m3 = (3/6) * total weight
Since the weights act vertically downward, the normal force acting on the pivot point must be equal in magnitude but opposite in direction to the vector sum of the weights. Therefore, the normal force can be calculated as follows:
N = Weight of m1 + Weight of m2 + Weight of m3
N = (1/6) * total weight + (2/6) * total weight + (3/6) * total weight
N = (6/6) * total weight
N = total weight
So, the normal force acting on the pivot point is equal to the total weight of the system.
(b) The ratio of L1 to L2 can be determined using the concept of center of mass. The center of mass of the system lies at the point of equilibrium, which is the pivot point. To find the ratio of L1 to L2, we need to consider the moments of the masses about the pivot point.
The moment of a mass about a point is given by the formula:
Moment = mass * distance from the point
For m1, the moment is m1 * L1 (distance from the pivot point).
For m2, the moment is m2 * L2 (distance from the pivot point).
Since the system is in equilibrium, the sum of the moments about the pivot point must be zero. Therefore, we can write the equation:
m1 * L1 + m2 * L2 = 0
To find the ratio L1/L2, we can rearrange the equation:
L1/L2 = -m2/m1
The negative sign indicates that L1 and L2 are in opposite directions from the pivot point. The ratio of L1 to L2 is equal to the ratio of the masses m2 to m1, but with opposite signs.
Note: The given information about the ratio of weights is not necessary to find the ratio of L1 to L2.