10. A ladder of mass M and length L situates on a frictionless floor and leans a

Question

10. A ladder of mass M and length L situates on a frictionless floor and leans against a frictionless wall. It is tied to the wall with a wire at the bottom as shown. There are four forces on the ladder, (1) the normal force upward from the frictionless floor (Fi) on the bottom of the ladder, (2) the horizontal tension force from the wire (T), also at the bottom of the ladder and to the right, (3) the downward Mg force, taken to be applied at the center of mass of the ladder, and (4) the normal force from the frictionless wall (Fw), pushing to the left at the very top of the ladder. Take the angle between the wire and the ladder as . For a rotation axis at the bottom of the ladder, which of the following torque balance equations is correct? (a) MgL(sint)/2=FL(cosa (c) MgL(sine/2-FwL(cost) (b) MgL(cos6)/2=FL(sina (d) MgL(cos6)/2-FwL(sin6)

Explanation / Answer

Mg is applied on the center of the ladder and Fw is applied on the tip of the ladder. Now, only those components of these force are to be calculated which are perpendicular to ladder. So, component of Mg is Mgcos? and that of Fw is Fwsin?. So, the torque balance equation is Mgcos? * L/2 = Fwsin? * L

(d) MgL(cos?)/2 = FwL(sin?)

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