Question

Two weights attached to a uniform beam of mass 58 kg are supported in a horizontal position by a pin and cable as shown in the figure.

Picture below

What is the tension in the cable which supports the beam? The acceleration of gravity is 9.8 m/s?
Answer in unite of kN

Answer 1

Answer:

1781.18 N, Counterclockwise

Explanation:

The problem depicts a beam supported by a pin and a cable, with two weights attached. To find the tension in the cable, we'll need to consider the torques about the pin (which we can assume to be the pivot point) and apply the principle that for the system to be in equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques.

$\boxed{ \left \begin{array}{ccc} \text{\underline{Torque:}} \\\\ \vec \tau = Fd\sin(\theta) \\\\ \text{Where:} \\ \bullet \ \vec \tau \ \text{is the torque} \\ \bullet \ F \ \text{is the force applied} \\ \bullet \ d \ \text{is the lever arm distance} \\ \bullet \ \theta \ \text{is the angle between the force vector and the lever arm} \end{array} \right.}$

We are given:

• m_b = 58 kg
• m₁ = 48 kg
• m₂ = 33 kg
• L = 4.1 m
• l = 2.7 m
• θ = 31°
• g = 9.8 m/s²

$\hrulefill$

Doing a sum of torques about point 'O':

$\sum \vec \tau_O: -\vec w_b\left(\dfrac{L}{2}\right)-\vec w_1l-\vec w_2L+\vec TL\sin(\theta) =0$

We know the formula for weight as:

$\boxed{ \left \begin{array}{ccc} \text{\underline{Weight of an Object:}} \\\\ \vec w = mg \\\\ \text{Where:} \\ \bullet \ \vec w \ \text{is the weight of the object (force due to gravity)} \\ \bullet \ m \ \text{is the mass of the object} \\ \bullet \ g \ \text{is the acceleration due to gravity} \end{array} \right.}$

Thus, our equation becomes:

$\Longrightarrow - m_bg\left(\dfrac{L}{2}\right)-m_1gl-m_2gL+\vec TL\sin(\theta) = 0$

Plug in our given values and solve for the tension in the cable 'T':

$\Longrightarrow -g\left(m_b\left(\dfrac{L}{2}\right)+m_1l+m_2L\right)+\vec TL\sin(\theta) = 0\\\\\\\\\Longrightarrow \vec TL\sin(\theta) = g\left(m_b\left(\dfrac{L}{2}\right)+m_1l+m_2L\right)\\\\\\\\\Longrightarrow \vec T = \dfrac{g\left(m_b\left(\dfrac{L}{2}\right)+m_1l+m_2L\right)}{L\sin(\theta)}$

$\Longrightarrow \vec T = \dfrac{(9.8 \text{ m/s}^2)\left((58 \text{ kg})\left(\dfrac{4.1 \text{ m}}{2}\right)+(48 \text{ kg})(2.7 \text{ m})+(33 \text{ kg})(4.1 \text{ m})\right)}{(4.1 \text{ m})\sin(31^\circ)}$

$\therefore \vec T \approx \boxed{1781.18 \text{ N}}$

Thus, the tension in the cable is approximately 1781.18 N. $\hrulefill$

Additional Information:

Determining the Direction of Torque:

In physics, the sign of torque (positive or negative) is determined by the direction of rotation it causes around an axis. If the torque tends to produce a counterclockwise rotation, it is considered positive. Conversely, if the torque tends to produce a clockwise rotation, it is considered negative. This convention is based on the right-hand rule.