A 560-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 560-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 140-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.) 0. ICE CKEAM (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. Indicate the direction with the sign of your answer) horizontal component vertical component

Explanation / Answer

NOTE - Take clockwise torque as negative and counter clockwise as positive

torque = force*distance

Now, taking torque about hinge point, we get

-140*3 - 560*4 + Tsin60*6 = 0 ------------- (1)

Here Tsin60 is the vertical component of tension T. and 60 degree is the angle between rod and wire and is calculated from triangle properties as 180 - ( 90+30) = 60

Solve (1) for T, we get

-420-2240 = -5.1962T

T = 511.9 Newtons

b) For horizontal component, we have

Fhorizontal = Tcos60 = 511.9cos60 = 255.95 N

For vertical

Fvertical + T sin60 = 140 + 560

Fvertical = 700 - Tsin60

Fvertical = 256.7 N.

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