The figure (Figure 1) shows a model of a crane that may be mounted on a truck. From the given information, the tension in the string A (TA) is [tex]\mathbf{ 202 \times 10^3 \ N}[/tex]
It is always a good idea to list the parameter given, this will help you highlight the known variables and assist you in finding the unknown variable. Here, we are to find the tension in string A.
From the given information;
- the mass of the rigid horizontal bar = 90.0 kg
- the length of the bar = 6.00 m
- the mass of the object attached at string B = 3500 kg
- the distance of the crane at B (x) = 5.80 m
- the magnitude of the acceleration due to gravity = 9.80 m/s²
The forces acting on the system can be seen in the image attached below.
The net torque applied on the bar is zero at equilibrium.
Using the application of Newton's Law for rotational motion, the net torque acting on the end of the bar can be computed as:
[tex]\mathbf{ \tau_{left} = 0}}[/tex]
[tex]\mathbf{{T_A (d) - W_1(\dfrac{L}{2}) -W_2(x) = 0}}}[/tex]
[tex]\mathbf{{T_A (d) - m_1g(\dfrac{L}{2}) -m_2g(x) = 0}}[/tex]
As such, the tension in string A can be determined as:
[tex]\mathbf{T_A(d) = m_1g (\dfrac{L}{2}) + m_2g (x)}[/tex]
[tex]\mathbf{T_A = \dfrac{m_1g (\dfrac{L}{2}) + m_2g (x)}{d}}[/tex]
[tex]\mathbf{T_A = \dfrac{(90.0 \ kg \times 9.80 \ m/s^2) (\dfrac{6.00 \ m}{2}) + (3500 \ kg )(9.80 \ m/s^2)(5.80 \ m)}{1.10 \ m}}[/tex]
[tex]\mathbf{T_A = 201586 \ N}[/tex]
[tex]\mathbf{T_A = 202 \times 10^3 \ N}[/tex]
Therefore, we can conclude that the tension in string A (TA) is [tex]\mathbf{ 202 \times 10^3 \ N}[/tex]
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