A uniform ladder 3 m long weighs 200 N is placed against a wall and floor as shown in the figure. The coefficient of friction between the wall and the ladder is 0.25 and that between the floor and ladder is 0.35. The ladder, in addition to its own weight, has to support a man of 1000 N at its top at B. Calculate: (i) The horizontal force P to be applied to ladder at the floor level to prevent slipping when the ladder making an angle of 60° with the floor. (ii) If the force P is not applied, what should be the minimum inclination of the ladder with the horizontal or floor, so that there is no slipping of it with the man at its top.
Man
Answers
Answer:
The free-body diagram for the ladder is shown below. We choose an axis through O, the top (where the ladder comes into contact with the wall), perpendicular to the plane of the figure, and take torques that would cause counterclockwise rotation as positive. The length of the ladder is L=10 m. Given that h=8.0 m, the horizontal distance to the wall is :
x=L2−h2=(10m)2−(8m)2=6.0 m
Note that the line of action of the applied force F intersects the wall at a height of (8.0m)/5=1.6m. In other words, the moment arm for the applied force (in terms of where we have chosen
the axis) is
r∣=(L−d)sinθ=(L−d)(h/L)=(8.0m)(8.0m/10.0m)=6.4m
The moment arm for the weight is x/2=3.0m, half the horizontal distance from the wall to the base of the ladder. Similarly, the moment arms for the x and y components of the force at the ground Fg are h=8.0 m andx=6.0 m, respectively. Thus, we have
Στz=Fr∣+W(x/2)Fg,xh−Fg,yx
=F(6.4m)+W(3.0m)+Fg,x(8.0m)−Fg,y(6.0m)=0
In addition, from balancing the vertical forces we find that W=Fg,y (keeping in mind that the wall has no friction). Therefore, the above equation can be written as
Στz=F(6.4m)+W(3.0m)+F
Explanation:
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Given : length of ladder = 3m
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.
Solution:
The horizontal force P to be applied to the ladder at the ground level to prevent slipping: The formula which can be used to solve is given as:-
Resolving forces horizontally,
P+Fs=Nw
Resolving forces vertically,
Ng+Fw=200+1000
Taking moments about A,
(1000 x 4cos 60°) + (200 x 2cos 60°)
(Fw x 3cos60) + (Nw x 3sin60)
2200 = 0.25 x Nw x 3cos 60° + Nw x 3sin 60°
Nw=739.97
Fw= 0.25 x 739.97
Fw= 184.99
Ng=1015.00
Fg= 0.35 x 1015.00
Fg= 355.25
P = 739.97-355.25
P = 384.72 N
The Force applied should be 384.72 N.