A 410-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a
ID: 1788202 • Letter: A
Question
A 410-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 155-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.)
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Tutorial Exercise A 410-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 155-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.00 angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.) 30.0° ICE CREAM SHOP (a) Find the (magnitude of the) tension T in the cable. (b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions.) Step 1 he free-body diagram of the sign-rod combination is shown belowExplanation / Answer
let,
weight of the sign,F1=410N
width, b=4m
height, h=3m
length of the rod , l=6m
weight of the rod, F2=155N
angle , theta=30 degrees
a)
by using equilibrium condition,
T*l*cos(theta)=F1*b+ F2*h
T*6*cos(30)=410*4+155*3
===> tension, T=405.1 N
b)
horizontal component, Fx=T*sin(30)
=405.1*sin(30)
=202.55 N
vetical component, Fy=F1+F2-Tcos(30)
=410+155-405*cos(30)
=214.26 N
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