Question

the free - body diagram of the sign - rod combination is shown below.
the tension in the cable and the reaction force exerted on the left end of the rod by the hinge have been resolved into horizontal and vertical components. the weight of the rod (vec{w}_{r}) acts at the center of the rod, ((6.00 m)/2 = 3.00 m) from the wall, and the weight of the sign (vec{w}_{s}) acts at the center of the sign, a total of (6.00 m-(4.00 m)/2 = 4.00 m) from the wall. note that the height of the sign does not figure into any of the calculations because it is not the lever arm of any torque and because we are given the weight of the sign.
consider a rotation axis perpendicular to the page, passing through the left end of the rod. we apply the second condition of equilibrium, (sum\tau = 0), to this system. we have
(+(tcos30.0^{circ})(6.00 m)-(155 n)(3.00 m)-(550 n)(4 m)=0)
or
(t=\frac{your response differs from the correct answer by more than 10%. double check your calculations.\times10^{3} ncdot m}{(6.00 m)cos30.0^{circ}} = 513e2 n)

Explanation:

Step1: Set up the torque equation

The torque equation for rotational equilibrium about the left - hand end of the rod is \(\sum\tau = 0\). The torque due to the tension \(T\) is \(\tau_T=(T\cos30^{\circ})(6.00\ m)\) (counter - clockwise torque), the torque due to the weight of the rod \(W_r\) is \(\tau_{W_r}=W_r(3.00\ m)\) (clockwise torque), and the torque due to the weight of the sign \(W_s\) is \(\tau_{W_s}=W_s(4.00\ m)\) (clockwise torque). So the equation is \((T\cos30^{\circ})(6.00\ m)-W_r(3.00\ m)-W_s(4.00\ m)=0\).

Step2: Solve for \(T\)

We can re - arrange the torque equation \((T\cos30^{\circ})(6.00\ m)=W_r(3.00\ m)+W_s(4.00\ m)\). Then \(T=\frac{W_r(3.00\ m)+W_s(4.00\ m)}{(6.00\ m)\cos30^{\circ}}\). Substituting \(W_r = 155\ N\) and \(W_s=550\ N\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(T\approx526\ N\)