To solve this problem, we need to analyze the forces and torques acting on the crane system. The crane consists of a long arm and a short arm, with a pulley block that can move along the long arm. Our goal is to find the minimum counterbalance required on the short arm to ensure that the system remains balanced and does not exceed the 10 percent imbalance limit.
Understanding the System
The crane has the following characteristics:
- Linear mass density of the long arm: 100 kg/m
- Length of the long arm: 50 m
- Length of the short arm: 5 m
- Weight of the pulley block: 500 kg
Calculating the Weight of the Long Arm
First, we need to determine the total weight of the long arm. Since the linear mass density is 100 kg/m and the length is 50 m, we can calculate the total mass:
Total mass of the long arm = Linear mass density × Length = 100 kg/m × 50 m = 5000 kg
Now, we can find the weight (force due to gravity) of the long arm:
Weight of the long arm = Total mass × g = 5000 kg × 9.81 m/s² = 49050 N
Finding the Torque Due to the Long Arm and Pulley Block
Next, we need to calculate the torques about the pivot point (where the crane is supported). The torque is given by the product of the force and the distance from the pivot point:
- Torque due to the long arm: This acts at the center of the long arm, which is at 25 m from the pivot.
- Torque due to the pulley block: This acts at a distance of 50 m from the pivot.
Calculating these torques:
Torque due to the long arm = Weight of the long arm × Distance from pivot = 49050 N × 25 m = 1226250 Nm
Torque due to the pulley block = Weight of the pulley block × Distance from pivot = (500 kg × 9.81 m/s²) × 50 m = 245250 N × 50 m = 1226250 Nm
Balancing the Torques
To maintain balance, the total torque on one side must equal the total torque on the other side. The counterbalance on the short arm will create a torque that must counteract the combined torque from the long arm and the pulley block.
Let’s denote the weight of the counterbalance as W_c. The torque due to the counterbalance is:
Torque due to counterbalance = W_c × 5 m
Setting the torques equal for balance:
Torque due to long arm + Torque due to pulley block = Torque due to counterbalance
1226250 Nm + 1226250 Nm = W_c × 5 m
2452500 Nm = W_c × 5 m
Solving for W_c gives:
W_c = 2452500 Nm / 5 m = 490500 N
Calculating the Mass of the Counterbalance
Now, we need to convert the weight of the counterbalance back to mass:
Weight = Mass × g, thus:
Mass = Weight / g = 490500 N / 9.81 m/s² ≈ 50000 kg
Considering the Imbalance Limit
The problem states that the vertical frame will twist and break if there is an excess imbalance of more than 10 percent. This means we need to ensure that the counterbalance does not exceed 10 percent of the total weight acting on the long arm and pulley block.
Total weight acting on the long arm and pulley block = Weight of long arm + Weight of pulley block = 49050 N + 4905 N = 53955 N
10 percent of this weight = 0.1 × 53955 N = 5395.5 N
Thus, the counterbalance must be less than or equal to 5395.5 N to avoid exceeding the imbalance limit.
Final Calculation of Minimum Counterbalance
To find the minimum counterbalance required, we can set the counterbalance weight equal to the maximum allowable imbalance:
W_c = 5395.5 N
Converting this back to mass:
Mass = 5395.5 N / 9.81 m/s² ≈ 550 kg
Therefore, the minimum counterbalance required on the short arm, which is to be installed on a permanent basis, is approximately 550 kg.
