A ladder can fall for two reasons. If it is set too steep and climber gets their

Question

A ladder can fall for two reasons. If it is set too steep and climber gets their mass to the left of the ladder's base, the ladder likely will fall over backwards. If the ladder is set at too shallow of an angle the required force of friction between the ladder and the ground might be too great and the base of the ladder will slip. Assume that there is no friction between the ladder and the wall and that the ladder is effectively weightless. The coefficient of friction between the base of the ladder and ground is 0.26. The person using the ladder will be 3/4 of the way up the ladder. If the person climbing the ladder has a weight of 980 newtons and the ladder is 4.95 meters long, how far from the wall can the base of the ladder be placed, to the nearest hundredth of a meter, and not slip?

Explanation / Answer

Now... vertical forces arebalanced: n = Mg . horizontal forces arebalanced: f = F . and torques are balanced. This means the sum of torques iszero. We will use the point where the ladder touches the ground asthe "rotation point" and measure the torques from there. . Torque from n? zero, because it acts at therotation point . Torque from friction? also zero, for the samereason . Torque from Mg? 0.75 L Mg sinf . because torque is force timesdistance from rotation point times sine of angle between them . Torque from F? - L F sin(90-f) . Total? 0.75 L M g sinf - L F sin(90-f) = 0 . or 0.75Mg sin f - F cosf =0 . or tanf = F / 0.75 Mg . Now you have threeequations: tanf = F / 0.75 Mg f =F n = Mg . You can use f =µn also for force offriction. . tanf = f / 0.75Mg = µn / 0.75 Mg = µMg / 0.75Mg = µ / 0.75 . tanf = 0.26 / 0.75 . f = 19.120 degrees. . Finally, distance from the wall can be found from the angle f,which is the angle the ladder makes with the vertical: . d = L sinf = 4.95 *sin19.120 = 1.62meters you're welcome. just in time hehe :)

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