A uniform horizontal bar of length L = 5 m and weight 219 N is pinned to a vertical wall and supported by a th?

A uniform horizontal bar of length L = 5 m and weight 219 N is pinned to a vertical wall and supported by a thin wire that makes an angle of theta = 25o with the horizontal. A mass M, with a weight of 322 N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 594 N.





What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks?





With M placed at this maximum distance what is the horizontal component of the force exerted on the bar by the pin at A?








With M placed at this maximum distance what is the vertical component of the force exerted on the bar by the pin at A?








http://i845.photobucket.com/albums/ab18/brittany_g18/prob10a.gif|||(a) We know that when M is placed at x, the system should be in equilibrium and moving M a distance dx from the pivot will cause the bar to fall. Meaning the sum of all forces and torques should be equal to 0. Let, m be the mass of the bar, since the bar is uniform, the centre of mass is at the geometric centre (L/2). Forces exerted by gravity are already perpendicular to the radius from the pivot (sin90=1), it is only necessary to take the sine of the angle for the tension (since it is 25 deg from the horizontal)





Pick the pivot point (from where torques will be calculated) to be at the point where the bar touches the wall


危蟿 = 0


mgL/2+Mgx=TLsin蠎


x=(TLsin蠎-mgL/2)/Mg





Now just plug and solve =D





(b) If you drew a free body diagram, you'd realize that the only horizontal force acting on the pin is F=Tcos蠎


because 危fx = 0





=P





(c)





危fy = 0


Mg+mg-Tsin蠎=F


F will be (+) meaning it will be directed downwards : )