A ladder of length 10 m and mass 20 kg (and
with uniform mass distribution leans against a
slippery vertical wall. The ladder makes an
angle of 30° with the vertical. Friction
between the ladder and the ground prevents it
from sliding downwards. What is the
5.
magnitude of the force exerted on the ladder
by the wall?
(Take V3 = 1.732 ; g = 10 m/s)
Answers
Answer:
There are four forces acting on the ladder of length L and making ?=53
degrees with the vertical smooth wall and on a horizontal rough floor.
1. the weight W=(10×9.8) = 98 N
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(
2
L
)sin?
3. Normal reaction N on the wall (horizontal) at the top end of the ladder,
equal in magnitude to
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(
2
L
)sin? = FLcos?
From which we can solve for F
= (
2
W
)tan?
= (
2
98
) tan 53
0
= 65 N
A 2.0kg block of wood is on a level surface Where v3 =0.81.732 and g=10m/s. A 58N force is being applied to the block parallel to the surface .(g=9.8m/s2). If the block was originally at rest