The diagram shows a bar with a length of 6.49 meters and a mass of 4.45 kilogram

Question

The diagram shows a bar with a length of 6.49 meters and a mass of 4.45 kilograms leaning against a frictionless wall. The bar makes an angle of 67.52 degrees with the floor. A block with mass 43.54 kilograms hangs from the bar a distance d up from the point of contact of the bar with the floor. As long as the distance, d, is not too great, the friction force exerted by the floor (s = 0.371) keeps the bar from slipping.

Find the maximum distance, d, that the hanging mass can be attached to the bar such that the bar does not slip.

Now the hanging mass is detached from the bar and removed from the problem. Find the minimum angle for which the bar will not slip.

Explanation / Answer

Normal force exerted on the bar by the floor. 43.54*9.8+4.45*9.8=470.3(N) Friction force F_fric=470.3*0.371=174.5(N). F_fric=N=174.5 (N) where N be normal force exerted by the wall. Net torque about contact point of the bar with the floor is zero. 174.5*6.49*sin67.52=4.45*9.8*6.49*cos(67.52)/2+43.54*9.8*d*cos(67.52) 1044.4=54.72+165*d. so d=6(m) ------------------------ the same as above. F_fric=m*g*k=N. so net torque is zero. N*L*sin(alpha)=mg*L*cos(alpha)/2 so tan(alpha)=mg/2N=mg/2*mgk=0.5/0.371 so alpha=53.4

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