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A heavy box is to be dragged along a rough horizontal floor . To do so ,person A pushes it an angle 30 0 from the horizontal and requires a minimum force

FA ,while person B pulls the box at an angle 60 0 from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is 3 1/2 / 5 ,the ratio FA/ FB is


The force of friction on the block in case 1 will be = uN The normal on the block will be reduced due to the vertical component of the FA . which is equal to - FAsin30 = FA/2. The net normal force on the block = (mg-FA/2) Therefore, the frictional force on the block = u(mg-FA/2). This frictional force will balance the horizontal component of the force that is - FAcos30 = FA3*(1/2)/2 Now, umg-uFA/2 = FA3*(1/2)/2 FA = umg/3*(1/2)/2+u/2 Similarly, we can find the value of FB using the angle as 60. And can find FA/FB by putting the value of coefficient o friction (u).
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