A uniform 12 kg 10 meters long rests against a vertical frictionless wall. The l

Question

A uniform 12 kg 10 meters long rests against a vertical frictionless wall. The ladder makes an angle theta of 73degree with the floor. A man 58 kg climbs slowly up the ladder. When he has climbed to a point that is 8.3 m that is 8.3 m from the base of the ladder, the ladder starts to slip. What is t e coefficient of static friction between the floor and the ladder? A rolling steel ball with mass m_1 = 0.38 kg moving to right 2.8 m/s collides elastically with second steel ball with mass m_2 = 0.22 kg moving to left 1.5 m/s. Find their velocities after the collision.

Explanation / Answer

5)
let mue_s is coefficint of static friction between laddr and floor.

before slipping, the ladder is in equilibrium.so, net torque and net force acting on it must be zero.

Normmal force acting on ladder, N = m*g + M*g

= 12*9.8 + 58*9.8

= 117.6 + 568.4

= 686 N

let Fw is the force exerted by the wall.

Apply, Fnetx = 0

mue_s*N - FW = 0

FW = mue_s*

= 686*mue_s

now Apply net torque about bootm point = 0

117.6*5*sin(17) + 568.4*8.3*sin(17) - FW*10*sin(73) = 0

117.6*5*sin(17) + 568.4*8.3*sin(17) = FW*10*sin(73)

117.6*5*sin(17) + 568.4*8.3*sin(17) = 686*mue_s*10*sin(73)

mue_s = (117.6*5*sin(17) + 568.4*8.3*sin(17))/(686*10*sin(73))

= 0.236 <<<<<<<<<------------Answer

6)

let

m1 = 0.38 kg, u1 = 2.8 m/s

m2 = 0.22 kg, u2 = -1.5 m/s

after the collsion,

v1 = (m1-m2)*u1/(m1+m2) + 2*m2*u2/(m1+m2)

= (0.38 - 0.22)*2.8 + 2*0.22*(-1.5)/(0.38+0.22)

= -0.652 m/s <<<<<<<<<------------Answer

v2 = (m2 - m1)*u2/(m1+m2) + 2*m1*u1/(m1+m2)

= (0.22 - 0.38)*(-1.5)/(0.38+0.22) + 2*0.38*2.8/(0.38+0.22)

= 3.94 m/s <<<<<<<<<------------Answer

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