A crane is configured as below, with the beam suspended at two points l1 and l2 by each end of a cable passing over a frictionless pulley. The two ends of the cable each make an angle θ with the beam. A counterbalance object C with mass mC is fixed at one end of the beam. A balance object B of mass mB is attached to the beam and can move horizontally in order to maintain static equilibrium. The crane lifts an object A with mass mA at a distance y from the counterbalance. For simplicity, assume the pulley, beam and cable to be massless.
(a) What is the tension in the cable that runs over the pulley? Express your answer in terms of mA, mA, mA, θ and acceleration due to gravity g (enter m_A for mA, m_B for mB, m_C for mC, theta for θ and g for g).
T=
(b) At what horizontal position, x, should one put the balance object B such that the crane doesn't tilt? Express your answer in terms of mA, mA, mA, l1, l2 and y (enter m_A for mA, m_B for mB, m_C for mC, l_1 for l1, l_2 for l2 and y for y).
x=
T=(m_A+m_B+m_C)/(2*sin(theta))*g
x=1/m_B*(((l_1+l_2)*(m_A+m_B+m_C))/2-y*m_A)
To find the tension in the cable that runs over the pulley, we can start by analyzing the forces acting on object A and the balance object B.
(a) The forces acting on object A are the weight (mg) directed downwards and the tension in the cable (T) directed upwards. Additionally, the tension in the cable also acts on the balance object B, but at different angles due to the different lengths of cable on either side of the pulley. Therefore, we can break down the tension into two components: T1 and T2, acting on the left and the right side of the pulley, respectively.
Since the system is in static equilibrium, the sum of the forces acting on both objects must be zero in both the horizontal and vertical directions.
In the horizontal direction:
The only force acting on object A is the horizontal component of T2, which is T2 * cos(θ). There is no horizontal force acting on object B.
Thus, T2 * cos(θ) = 0
T2 = 0
In the vertical direction:
The total vertical force on object A is the weight of object A (mA * g) minus the vertical component of T1.
The total vertical force on object B is the vertical component of T1 plus the weight of object B (mB * g) plus the weight of the counterbalance object C (mC * g).
Therefore, we have:
T1 * sin(θ) = mA * g - mB * g - mC * g
T1 = (mA - mB - mC) * g / sin(θ)
So, the tension in the cable that runs over the pulley is T = T1 = (mA - mB - mC) * g / sin(θ).
(b) To find the horizontal position x at which the crane doesn't tilt, we need to consider the torques acting on the system.
The torque exerted on the system by object A is the weight of object A (mA * g) multiplied by its perpendicular distance to the pivot point (which is y). This torque tends to rotate the system in a counterclockwise direction.
The torque exerted by the balance object B is the weight of object B (mB * g) multiplied by its perpendicular distance to the pivot point (which is l2 - x). This torque tends to rotate the system in a clockwise direction.
The torque exerted by the counterbalance object C is the weight of object C (mC * g) multiplied by its perpendicular distance to the pivot point (which is l1).
For the system to be in static equilibrium and not tilt, the sum of the torques must be zero.
Therefore, we have:
(mA * g * y) - (mB * g * (l2 - x)) - (mC * g * l1) = 0
Simplifying this equation, we get:
mA * g * y - mB * g * l2 + mB * g * x - mC * g * l1 = 0
Rearranging, we can solve for x to find the horizontal position at which the crane doesn't tilt:
x = (mA * g * y - mC * g * l1) / (mB * g) + l2
So, the horizontal position x should be (mA * g * y - mC * g * l1) / (mB * g) + l2.