Answer:
Force of friction = 1926.18NN
Step-by-step explanation:
There are four forces acting on the ladder of length L and making ?=50°
degrees with the vertical smooth wall and on a horizontal floor.
Total weight = weight of ladder + weight of worker = 25 + 85 = 110kg
1. the weight W=(25 + 85) × 9.8 = 1078N
acting downwards through the middle of the ladder
2. vertical reaction on the floor acting upwards, equal to W.
These two forces form a couple of magnitude
W(2L)sin?
3. Normal reaction N on the wall (horizontal) at the top end of the ladder,
4. Frictional force F acting horizontally at the bottom of the ladder.
These two forces form another couple equal to FLcos?.
Since the ladder is in equilibrium, the two couples must be equal, thus
W(L/2)sin? = FLcos?
The man is standing 2/3 of the length of the ladder, so we divide by 2/3
From which we can solve for F
F = (W/0.667)Tan 50°
F= (1078/0.667 ) × 1.1918
F = 1616.19 × 1.1918
F = 1926.18N