A uniform beam of length 1.2 m and mass 18 kg is attached to a wall by a cable t

Question

A uniform beam of length 1.2 m and mass 18 kg is attached to a wall by a cable that makes an angle of 42 degrees with the end of the beam, as shown in the figure. The beam is free to pivot at the point where it attaches to the wall. The figure should is below.

(a) What is the torque on the beam from the force of gravity on it? Include the magnitude and the sign (positive or negative) according to the standard convention.

(b) What is the tension in the cable? (Magnitude and sign)

(c) What are the magnitude and direction of the force that the pivot mechanism applies to the beam?

Explanation / Answer

Torque = Fd = 18*9.8* 0.6 = -105.84
Tsin42*1.2 = 105.84 hence T = +131.8

pivot Ry = W- Tsin42 = 18*9.8 - 131.8sin42 = +88.2
Rx = Tcos42 = +97.95

magnitude of reaction force = (Rx^2 + Ry^2) = 131.81 N

direction = arctan(Ry/Rx) = 42

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.