A 520 N uniform rectangular sign 4.00m wide and 3.00 m high is suspended from a

Question

A 520 N uniform rectangular sign 4.00m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long,uniform, 100-N rod as indicated in thefigure below. The left end the rod is supported by a hinge and theright end is supported by a thin cable making a 30.0° anglewith the vertical.


(a) Find the tension, T, in the cable.
1 N
(b) Find the horizontal and vertical components of force exerted onthe left end of the rod by the hinge. (Take up and to the right tobe the positive directions.) horizontal component 2 N vertical component 3 N horizontal component 2 N vertical component 3 N

Explanation / Answer

first, write horizontal and vertical forces .     horiz forces arebalanced:    right forces = left forces .                              Wx   =    T sin30 .    vertical forces are balanced:   up forces = down forces .                         T cos30 + Wy   =   520 +100 . You have two equations and threeunknowns     T Wx  Wy      (these are the threethings you have to find). . Now you can do torques on the rod, using the left end as thereference point: .           uptorques = down torques .           torque from cable = torque from weight of sign + torquefrom weight of rod .            T * 6.00* sin60   = 520 *4.00   +   100 * 3.00 . Solve for T .          T = (2080 + 300) / 6.00 sin60 =     458 N . Now get the other two .     Wx   = 458 *sin30 =    229 N .    Wy = 620 - 458*cos30 =    223N

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