The figure shows a 4.20-kg, 1.80-m-long rod hinged to a vertical wall and supported by a thin wire. The wire and rod each makeangles of 45° with the vertical. When a 10.0-kg block is suspended from the midpoint of the rod, the tension T in the supportingwire is 49.3 N. The wire will break when the tension exceeds 75.0 N.45°45°T10kgTipler & Mosca, Physics for Scientists andEngineers, 6e © 2008 W.H. Freeman andCompanyWhat is the maximum distance dmax from the hinge fromwhich the block can be suspended?dmax0.871Incorrectm

Given Data: The mass of the rod, M = 4.20 kg. The length of the rod, L = 1.80 m. The wire and the rod each make angles of θ = 45° with the vertical. The mass of the block that is suspended from the midpoint of the rod is , m = 10 kg. The maximum limit of tension for the wire is T = 75.0 N. To Find: The maximum distance dmax Solution: Here we will use the second condition of equilibrium to solve this problem. Let, assume the maximum distance be x. The second condition of equilibrium states that the net torque acting on the object must be zero. So ,according to the second condition of equilibrium ,we can write Σ J =0i.e TL -MgL2sin 45°-mgxsin 45°=0 Solving for x we get, g=acceleration due to gravity g=9.81ms2 x=TL-MgL sin45°2mg Sin45° Now , putting all the known values , we get x=75×1.80-4.20 ×1.80×9.81×sin45°210×9.81×sin45°x=1.58681m The maximum distance dmax from the hinge from which the block can be suspended , dmax =1.5681 mAnswer: Therefore,the maximum distance dmax from the hinge from which the block can be suspended , dmax =1.5681 m

Answered: The figure shows a 4.20-kg, 1.80-m-long…