Answer to Question #238927 in Mechanical Engineering for Vasu

Question

A 4 m long ladder is supported by a wall and floor as shown in Figure. The coefficient of friction at the wall is 0.30 and that the floor is 0.50. The weight of the ladder is 200 N. The ladder supports a vertical load of 800 N at C. Solve for the reactions at the floor and wall. Compute the least value of at which the ladder may be placed without slipping.

Answer 1

length of ladder = 4m

Weight of ladder = 200 N

∠ made by ladder against the wall = 60°

To Find : (i) The horizontal force P to be applied to the ladder at the ground level to prevent slipping.

(ii) If the force P is not applied, what should be the minimum inclination of the ladder with the horizontal, so that it does not slip with the man at the top

Step-by-step explanation:

i) The horizontal force P to be applied to the ladder at the ground level to prevent slipping:

Resolving forces horizontally,

P + F_{8} = N_{w}P+F

8

​

=N

w

​

Resolving forces vertically,

N_{g} + F_{w} = 200 + 1000N

g

​

+F

w

​

=200+1000

Taking moments about A,

(1000 x 4cos 60°) + (200 x 2cos 60°)

(F_{w}F

w

​

x 4cos 60°) + (N_{w}N

w

​

x 4sin 60°)

2200 = 0.25 x N_{w}N

w

​

x 4cos 60° + N_{w}N

w

​

x 4sin 60°

N_{w}N

w

​

= 554.98 N

F_{w}F

w

​

= 0.25 x 554.98

= 138.75 N

N_{g}N

g

​

= 1061.25 N

F_{g}F

g

​

= 0.35 x 1061.25

= 371.44 N

P = 554.98 - 371.44

= 183.54 N