Figure (a) shows a vertical uniform beam of length L that is hinged at its lower

Question

Figure (a) shows a vertical uniform beam of length L that is hinged at its lower end. A horizontal force Fa 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 ? with the horizontal. Figure (b) 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 Ts = 650 N. Figure (c) gives the magnitude Fh of the horizontal force on the beam from the hinge, also as a function of y/L. Evaluate (a) angle ? and (b) the magnitude of Fa.

Explanation / Answer

We use Newton's second law, x axis

Fa-T Cos - Fh = 0

Second law of Newton rotational Torque

T Cos L - Fa y = 0 + signe spin counterclockwise

We write the set of equations

-T Cos =- Fa + Fh

T Cos = Fa y/L

We solve by adding

0 = Fa (y/L - 1) + Fh

Fa = - Fh / (y/L - 1)   

We are looking for the equation of a straight line with each graph

. T = m (y/L) + b

Points T (1) = 0 y/L (1) = 0

(2) = 650 N t y/L (2) = 1

Slope m = (650-0) /(1-0) = 650 N

T = 650 (y/L)

. Fh = m (y/L) + b

Points Fh(1) = 270 + 67 Fh 5 N y/L (1) = 0

Fh (2) = 0 y/L (2) = 1 m = (0 - 337.5) / (1-0) = - 337.5 N

Fh = - 337.5 (y/L) + 337.5

We are looking for the applied force Fa

Fa = - (-337.5y/L + 337.5)) /(y/L-1) = (337.5 (y/L) - 337.5) / (y/L-1) = 337.5 (y/L-1) /(y/L-1)

Fa = 337.5 N

We are looking for the angle, entering in the torque equation

T L Cos = Fa y Cos = Fa y / (L T) = f (y/L) /T

Cos = 337.5 (y/L) / (650 y/L)

COS = 337.5 650 = 0.519 = 58 °

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