ˆˆ( 45 N)i+(200 N)j.gF=−G(c) Note that the phrase “start to move towards the wall” implies that the friction force is pointed away from the wall (in the −#idirection). Now, if f = –Fgxand FN= Fgy= 200 N are related by the (maximum) static friction relation (f = fs,max= μsFN) with s= 0.38, then we find Fgx= –76 N. Returning this to the above equation, we obtain 2(200 N)(3.0m) (76 N)(8.0m)1.9 10 N.6.4mF+==×85. We choose an axis through the top (where the ladder comes into contact with the wall), perpendicular to the plane of the figure and take torques that would cause counterclockwise rotation as positive. Note that the line of action of the applied force GFintersects the wall at a height of (8.0 m)/5 1.6m=; in other words, the moment armfor the applied force (in terms of where we have chosen the axis) is (4/5)(8.0 m)r⊥. The moment arm for the weight is half the horizontal distance
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.