Question

A 100.-kg uniform bar of length \(L=5.00 \mathrm{~m}\) is attached to a wall by a hinge at point \(A\) and supported in a horizontal position by a light cable attached to its other end. The cable is attached to the wall at point \(B\), at a distance \(D=2.00 \mathrm{~m}\) above point \(A\). Find: a) the tension, \(T,\) on the cable and \(b\) ) the horizontal and vertical components of the force acting on the bar at point \(A\)

Short answer

To find these values, follow these steps: 1. Draw a free-body diagram of the bar, labeling the weight (W), tension (T), and force components at the hinge (F_horizontal and F_vertical). 2. Resolve T into horizontal (T_x) and vertical (T_y) components using trigonometry and the angle θ between the cable and the horizontal. 3. Calculate the torques around point A created by the weight of the bar (τ_W) and the vertical component of tension (τ_Ty). 4. Equate the torques and solve for tension T. 5. Calculate F_horizontal and F_vertical using the equilibrium condition and the calculated tension. By doing this, we can determine the tension in the cable and the horizontal and vertical components of the force acting on the bar at point A.

Step-by-step solution

Draw a free-body diagram of the bar

Draw a diagram of the bar with the hinge at point A and the cable attached to point B. Label the weight of the 100 kg bar as \(W = mg\) (where \(m = 100 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\)) acting downward at the center of the bar, the tension (T) in the cable acting upward and rightward from the bar's right end, and the horizontal (F_horizontal) and vertical (F_vertical) forces at the hinge.

Resolve T into horizontal and vertical components

Using the geometry of the problem, find the angle θ below the horizontal at which the cable pulls on the bar. In this case, the angle θ can be calculated as: \(\theta_{T} = \tan^{-1}(\frac{D}{L})\) Substitute the values given for D and L to determine the angle: \(\theta_{T} = \tan^{-1}(\frac{2}{5})\) Now, resolve tension T into horizontal and vertical components using trigonometry: \(T_x = T \cos{\theta_{T}}\) \(T_y = T \sin{\theta_{T}}\)

Calculate the torques around point A

Calculate the torque created by the weight of the bar \(W\) and by the vertical component of tension in the cable \(T_y\). The torque about point A can be calculated as follows: \(\tau = F \times d \times \sin{\alpha}\) Where \(F\) is the force, \(d\) is the distance from the pivot point, and \(\alpha\) is the angle between the force and the distance vector. For the torque created by the weight: \(\tau_W = W \times \frac{L}{2} \times \sin{90^{\circ}} = W \times \frac{L}{2}\) For the torque created by the vertical tension component: \(\tau_{T_y} = T_y \times L \times \sin{90^{\circ}} = T_y \times L\) Since the bar is in equilibrium, the torques are equal and opposite. Therefore, \(\tau_W = \tau_{T_y}\). Substitute the expressions for torque and solve for tension T.

Calculate the tension T

Using the equal torques and substituting the expression for \(T_y\), we have: \(W \times \frac{L}{2} = T_y \times L = (T \sin{\theta_{T}}) \times L\) Divide both sides by \(\sin{\theta_{T}}\) and L: \(T = W \times \frac{1}{2} \times \frac{1}{\sin{\theta_{T}}}\) Substitute the values for W, L, and the calculated angle \(\theta_{T}\) and solve for the tension T.

Calculate the horizontal and vertical forces at point A

We now have enough information to determine the horizontal and vertical components of the force acting at point A. To do this, we just need to apply the following equations using the equilibrium condition: \(\Sigma F_x = F_{horizontal} - T_x = 0\Rightarrow F_{horizontal} = T_x = T \cos{\theta_{T}}\) \(\Sigma F_y = F_{vertical} - W + T_y = 0\Rightarrow F_{vertical} = W - T_y = W - (T \sin{\theta_{T}})\) Using the tension calculated earlier and the angle θ, plug in the values to compute the horizontal and vertical components of the force acting on the bar at point A.

Key concepts

Torque in Static Equilibrium

Torque plays a pivotal role in static equilibrium problems, as it represents the rotational effect of forces acting on an object. In the context of a bar held horizontally by a hinge and a cable, torque is essential for understanding how the bar remains stationary despite the forces acting on it.

Consider the bar as a see-saw with the hinge at point A being the fulcrum. To achieve equilibrium, the torques created by the weight of the bar and the tension in the cable must balance each other. Mathematically, torque (\tau) is calculated as the product of force (F), distance (d) from the pivot to the point where the force is applied, and the sine of the angle (\theta) between the force and the distance vector. The equation is represented as \(\tau = F \times d \times \sin{\theta}\).

In this case, the force due to gravity acts downward at the center of the bar, while the tension's vertical component acts upward at the end of the bar. The distances from point A to these points of force application and the angles they make with the horizontal are crucial in determining torque. When the bar is in equilibrium, these torques are equal, allowing for the calculation of unknown forces such as the tension in the cable.

Understanding the Free-Body Diagram

A free-body diagram is a visual representation of the forces acting on an object. It is a crucial tool for solving problems in statics as it helps to identify all the relevant forces and how they interact.

To draw a free-body diagram for the bar, we start by considering all the points where forces are applied. The hinge at point A provides a pivot, while the tension T acts on the other end of the bar. The weight of the bar acts at the geometric center. Gravity's force is represented by a vector pointing downward, and the tension's vector is at an angle to the horizontal, as specified by trigonometry given the points of attachment. The forces at the hinge are split into horizontal and vertical components, but their exact magnitudes only become apparent after considering the equilibrium of torques and the forces at the other end of the bar.

By carefully labeling these forces and breaking them down into components, the free-body diagram lays out a roadmap for solving for unknown quantities, such as the tension in the cable and the forces at the hinge, based on static equilibrium conditions.

Calculating Tension in a Cable

Tension is the force transmitted through a cable when it is pulled tight by forces acting at either end. In the exercise involving the horizontal bar, the cable's tension must counter the moment caused by the bar's weight. To maintain static equilibrium, the tension provides just enough force to keep the bar from rotating.

The tension T can be decomposed into horizontal (\(T_x\)) and vertical (\(T_y\)) components. These components are derived using trigonometric functions, specifically the sine and cosine of the angle at which the cable is attached to the wall. This angle, in turn, depends on the geometry of the system—how high above the hinge the cable is fixed (\(D\)) and the length of the bar (\(L\)).

To find the tension's vertical component, use \(T_y = T \sin{\theta_{T}}\), which directly contributes to the equilibrium condition for torques (\(\tau\)). Since the sum of the torques must equal zero in static equilibrium, the product of tension's vertical component and the bar's length (\(T_y \times L\)) will be set equal to the torque due to the bar's weight (\(W \times \frac{L}{2}\)). The actual value of T can then be determined from \(T_y\) using trigonometric relationships, ultimately completing the analysis of tension within the system.