Question

ab is a uniform rod with length 60 cm. and weight 40 newton is connected to a hinge on the vertical wall at a. if the rod keep t in equilibrium horizontally by a light string connected to the rod at b and with point c on the wall just above a and at a distance 60 cm. from a. find the magnitude of the tension on the string and the magnitude of the reaction on the hinge at a.

Explanation:

Step1: Calculate the torque due to the weight of the rod

The weight of the rod \(W = 40\space N\) acts at the mid - point of the rod (since it is uniform). The length of the rod \(L=60\space cm = 0.6\space m\). The torque due to the weight about point \(A\), \(\tau_{W}\) is given by \(\tau_{W}=W\times\frac{L}{2}\).

$$ \tau_{W}=40\times\frac{0.6}{2}=12\space N\cdot m $$

Step2: Calculate the torque due to the tension in the string

Let the tension in the string be \(T\). The distance from \(A\) to the point of application of tension (assuming the rod is horizontal and the string is vertical - like in a typical lever problem, if we consider the geometry of the rod and the string connection). Let the length of the rod \(L = 0.6\space m\) and the distance from \(A\) to the point of application of tension (assuming the string is attached at the end \(B\) - a common case for a rod - string - hinge system). The torque due to tension \(\tau_{T}\) about point \(A\) is \(\tau_{T}=T\times L\) (using the formula \(\tau = rF\sin\theta\), and if the string is perpendicular to the rod \(\theta = 90^{\circ},\sin\theta = 1\)).
Since the rod is in equilibrium, \(\tau_{W}=\tau_{T}\) (sum of torques about the hinge \(A\) is zero, \(\sum\tau_{A}=0\)).

$$ T\times0.6 = 12 $$

$$ T=\frac{12}{0.6}=20\space N $$

Step3: Calculate the reaction at the hinge

For vertical equilibrium (\(\sum F_{y}=0\)), let the vertical component of the hinge reaction be \(A_{y}\) and for horizontal equilibrium (\(\sum F_{x} = 0\), assume no horizontal forces other than the hinge reaction \(A_{x}\), if the string is vertical). Since the rod is in equilibrium \(\sum F_{y}=0\), \(A_{y}=W - T\) (if we assume the positive \(y\) - direction is upwards). \(A_{y}=40 - 20=20\space N\) and \(A_{x} = 0\) (if no horizontal forces from the string). The magnitude of the hinge reaction \(R=\sqrt{A_{x}^{2}+A_{y}^{2}}\)

$$ R = 20\space N $$

Answer:

The magnitude of the tension in the string is \(20\space N\) and the magnitude of the reaction at the hinge \(A\) is \(20\space N\)