A uniform ladder 5.0 m long rests against a frictionless, verticalwall with its

Question

A uniform ladder 5.0 m long rests against a frictionless, verticalwall with its lower end 3.0 m from the wall. The ladder weighs160 N. The coefficient of static frictionbetween the foot of the ladder and the ground is 0.40. A manweighing 710 N climbs slowly up theladder. (a) What is the maximum frictional force thatthe ground can exert on the ladder at its lower end?
1 N

(b) What is the actual frictional force when the man has climbed1.0 m along the ladder?
2 N

(c) How far along the ladder can the man climb before the ladderstarts to slip?
3 m (a) What is the maximum frictional force thatthe ground can exert on the ladder at its lower end?
1 N

(b) What is the actual frictional force when the man has climbed1.0 m along the ladder?
2 N

(c) How far along the ladder can the man climb before the ladderstarts to slip?
3 m

Explanation / Answer

(a) max friction = u * normal = u * total weight = 0.40 * (160+710) =    348Newtons . (b) The ladder creates a   3-4-5 triangle with the wall and floor. The angle between the ladder andvertical is .           arcsin (3/5) =    36.87 degrees      and with thevertical,     53.13 degrees . Now consider the torques acting on the ladder, using the pointwhere the ladder meets the wall as reference: .     torque from weight of man + torque from weight of ladder + torque from friction = torquefrom normal .           710 *4 * sin53.13    +   160 * 2.5 *sin53.13   +   f * 5 * sin36.87  = 870 * 5 * sin53.13 .        710*4*0.8    +  160*2.5*0.8    +  f*5*0.6    =   870*5*0.8 . Solve for f...                          f =    296 Newtons    isthe actual friction force . (c) now we use the same torque equation, but set f equal to the max possible and solve for the distance theman can walk up the ladder: .          710 * d *0.8   +   160 * 2.5 *0.8     +   348 * 5 * 0.6  = 870 * 5 * 0.8 . Solve ford...          d = 3.725 meters

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