Figure 12-49ashows a vertical uniform beam of length Lthat is hinged at its lower end. A horizontal forceFa is applied tothe beam at distance yfrom the lower end. The beam remainsvertical because of a cable attached at the upper end, at angleθwith the horizontal. Figure12-49agives the tension Tin the cableas a function of the position of the applied force given as a fraction y/Lof the beam length. The scale of the Taxis is set byTs=600N.Figuregives the magnitude Fhof the horizontal force on thebeam from the hinge, also as a function of y/L. Evaluate (a) angleθand (b) the magnitude of Fa.

Short Answer

Expert verified
  1. Angleθis60
  2. Magnitude ofFa is 300 N

Step by step solution

01

Determine the given quantities

The scale of the Taxis is set by Ts=600N.

02

Determine the concept and the formulas:

If net torque acting on the system is zero, the object is said to be in equilibrium. This can happen when there are more than one torque acting on the object in a different direction such that the vector sum of all the torques acting on the object is zero. As per Newton’s first law, if there is no torque acting on the system, the object will continue its state of rest or state of angular motion.

Formula:

TLcosθ-Fay=0

03

(a) To calculate angle

With the pivot at the hinge, the net torque zero. It can be written as,

TLcosθ-Fay=0

This leads to,

T=FacosθyL …… (ii)

It can be interpretedFa/cosθ as the slope on the tension graph. To getFh use,

Fh=Tcosθ-Fa=-FayL-Fa

Substitute equation (ii), the result implies that the slope on the Fh graph is equal to -fa or fa=300. Thus substituting this value we get,

θ=60

04

(b) Calculate the magnitude of Fa→

The magnitude of the slope of Fh graph is equal to the magnitude of Fa.

Consider point A and point B on Fh graph as shown in figure below.


The slope is calculated as,

Slope=By-AyBx-Ax=0-1201-0.6=-1200.4=-300

Thus, the magnitude ofFa=300N

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