Answer to Question #266754 in Physics for Frizie

Question #266754

1. A 3.0 meters long solid shaft makes an angle of 35 degrees with the horizontal. A vertical


force of 50 N is applied 0.60 m from the upper end. What is the magnitude of the torque


due to the vertical force about each end?


2. A Ladder of 6.0 meters in length weighs 150 N rests on horizontal ground and leans at an


angle of 65 degrees with the horizontal against a smooth vertical wall. How far up the ladder


may a 700 N person go before the ladder slips? The coefficient of friction between ladder


and ground is 0.40.


3. A Flat bar is loaded with boxes as shown in


figure at right. Box M = 50 kg, what must be


the magnitude of mass m so that the bar


becomes level? Neglect the weight of the bar.


What torque about each end of the bar?

1
Expert's answer
2021-11-16T19:20:28-0500

1) The torques from the vertical force around the upper end is the lever times vertical component of the force:


"\u03c4 _\nu\n\u200b\n =\u2212F \\cos\u03b8\u22c5l=50\\cos35\u00b0\u22c50.6=\u221224.6\\ Nm."

Around the lower end:


"\u03c4 _\nl\n\u200b\n =F \\cos\u03b8\u22c5(L\u2212l)=50 \\cos35\u00b0(3\u22120.6)\\\\=\n98.3\\ Nm."

3) If the length of the bar is L, and its axis of rotation is at distance d from the point where mass M is attached, the value of mass m in order to balance M can be found from the equilibrium of torques:


"Mgd=mg(L\u2212d),"

The torque about each end of the bar is


"\u03c4 _\nM\n\u200b\n =Md,\\\\\n\u03c4 _\nm\n\u200b\n =m(L\u2212d)."


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