A uniform ladder having a mass of 50 kg is placed with its top against a rough vertical wall, and its foot on a horizontal ground. Find the minimum mass P kg of the block attached to the foot of the ladder to retain it at an angle of 45°. Coefficient of static friction between the ladder and the wall and the floor and also between the block and the floor, μ = 0.20.
Answer 237.
The forces acting on the ladder when it is on the verge of sliding are
1) Its weight W = mg = 50 *9.81 = 490.5 N
2) Normal reaction from the floor, R1, upwards
3) Normal reaction from the wall, R2, horizontal in the direction of sliding of the ladder
4) Frictional force (0.20)R1 at the floor, opposite to the direction of sliding
5) Frictional force (0.20)R2 at the wall upwards
6) Frictional force due to mass P = 9.81 * 0.20 P in the direction opposite to the motion.
Taking moments of these forces at the point of contact with the floor,
50 * 9.81 * (1/2) = 1.20 R2
=> R2 = 204.4 N
Balancing vertical forces,
R1 = 50 * 9.81 - (0.20) * (204.4) = 449.6 N
Taking moments of all the forces about the point of contact of the ladder with the wall,
50*9.81*(1/2) + 0.20 * (449.6) + P*9.81*0.20 = 449.6
=> 1.962 P = 449.6 - 245.25 - 89.9
=> 1.962 P = 114.45
=> P = 114.45/1.962 = 58.3 N.
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