Question

The figure shows a vertical uniform beam of length L that is hinged at its lower end. A horizontal force is applied to the beam at distance y from the lower end. The beam remains vertical because of a cable attached at the upper end, at angle u with the horizontal. The figure gives the tension T in the cable as a function of the position of the applied force given as a fraction y/L of the beam length. The scale of the T axis is set by $$T_s = 600$$ N. The figure gives the magnitude $$F_h$$ of the horizontal force on the beam from the hinge, also as a function of y/L. Evaluate (a) angle u and (b) the magnitude of $$\vec F_a$$.

Answer 1

(a) With the pivot at the hinge, Eq. 12-9 yields
$$TL\cos\theta-F_ay=0$$
This leads to $$T = (F_a/\cos\theta)(y/L)$$ so that we can interpret $$F_a/\cos\theta$$ as the slope on the tension graph (which we estimate to be 600 in SI units). Regarding the $$F_h$$ graph, we use Eq. to get
$$F_h = T\cos\theta − F_a = (−F_a)(y/L) − F_a$$
After substituting our previous expression, the result implies that the slope on the $$F_h$$ graph (which we estimate to be –300) is equal to $$−F_a$$ , or $$F_a = 300$$ N and (plugging back in) $$\theta = 60.0^\circ$$.
(b) As mentioned in the previous part, $$F_a = 300$$ N.