Scie 1001
Question 1
Question 2
Question 3
Question 4



Questions and Answers


1. A uniform ladder weighing 200N leans against a wall as in Fig 1. The forces on the ladder are W (vertically downwards), Fw (horizontal) and Fg (at an angle of 60° to the horizontal).
(a) Consider there is no friction between the wall and the top of the ladder, draw a diagram to represent the forces on the ladder.
(b) Calculate the friction force Fh (horizontal force on the foot of the ladder) from the floor.
(c) Calculate the normal reaction Fv (vertical force on the foot of the ladder) from the floor.
(d) Calculate the resultant force from the ground Fg and its angle, θ, to the vertical.
(e) Calculate Fw (the reaction from the wall).

Fig 1


(b) The force Fg can be resolved into two forces at right angles to each other Fh and Fv.

Recall Sin θ = opposite/hypotenuse
and Cos θ = adjacent/hypotenuse

so Sin 60° = Fv/Fg -----Eq 1

and Cos 60° = Fh/Fg --- Eq2

From Eq1 Fg = = Fh/Cos 60° --- Eq3

From Eq2 Fg = Fv/Sin 60° --- Eq4
Fig 2  
Figure 2 can be redone to show the components of Fg (Fv and Fh)

The ladder is in equilibrium:
Recall from Newton's First and Second Laws the conditions for static equilibrium:

∑F = 0
∑Horizontal forces = 0 --- Eq5
∑Vertical forces = 0
--- Eq6

Fig 3

(c) From Eq6
Fv + W = 0
Fv = -W
Taking up as positive
Fv = -(-200N)
Fv = 200N up
(d) From Eq4
Fg = = Fv/Sin 60°

Fg = 200N/Sin 60°

Angle between Fg and the vertical
90° - 60° = 30°
Fg = 231N at 30° to the vertical

(b) From Eq3
Fg = Fh/Cos 60°

Fg Cos 60° = Fh
Fh = 115N

(e) From Eq5
Fh + Fw = 0
Fh = -Fw

Taking to the right as positive
-Fw = Fh
-Fw = - 115N
Fw =115N


Double Checks for the answers Fg, Fv and Fh.
By Pythagoras:

Fh2 + Fv2 = Fg2
1152 + 2002 = 2312










(b) Fh = 115N (to the left)
(c) Fv = 200N (up)
(d) Fg = 231N (at 30° to the vertical)
(e) Fw =115N (to the right)
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.