Assume the beam is uniform. The free-body diagram is shown below, wher w b =m b

Question

Assume the beam is uniform. The free-body diagram is shown below, wher wb =mbg, and wp = mpg. +counterclockwise at A = 0;(Tsin)(L) - (Wb)(L/2) - (Wp)(x) =0...(1). +counterclockwise at A = 0;(Tsin)(L) - (Wb)(L/2) - (Wp)(x) =0...(1). Given mb + mp = 62.30kg...(2). When x = 0, T = 500N. When x = L, T = 700N. When x = L/2, T = 600N. Putting x = L/2 into (1), we get (Tsin)(L) =(mb + mp)(g)(L/2) Putting x = 0 into (1), we get (Tsin)(L) =(mb)(g)(L/2). Solve for mb =2(500N)(sin30.620)/(9.81m/s2) = 51.92kg.

Part C is to find mp. I tired plugging into formula, but the answercomes out wrong.

Explanation / Answer

You are doing it correctly , As in the question , it is given that mb + mp= 62.30kg mb=51.92 kg So mp = 62.30 - 51.92 = 10.38 Also using the third condition ,i.e. when x= L , T = 700 N Putting in the equation , (Tsin)(L) - (Wb)(L/2) - (Wp)(x) = 0 700*sin 30.60*L -51.92*9.81*L/2 - Wp*L =0 356.3 - 254.67 -mp*9.81 =0 mp = 10.37 We can see that we get the same value using two equation So mp should be 10.38 kg

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