True or false. Lateral to the inguinal region is the coxal region, while superior to the scapular region is the dorsal region.
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Answer:
Check attachment for the free body diagram
Explanation:
Given that,
Ladder length L = 14m
Weight of ladder W= 490N
The weight will act at the midpoint
i.e at 14/2 = 7m, L1 = 7m
The ladder makes an angle of 55° with the horizontal. θ=55°
Weight of firefighter Wf =810N
The firefighter is at 3.9m from the horizontal ground, L2 =3.9
The wall exerts a force on the ladder, let It be Nw
The ground exerts a force on the ladder, let it be Ng
The let Ff be the frictional force that opposes motion.
a. We want to find the horizontal and vertical force the ground exerted on the ladder i.e Ng and Ff.
Using Newton second law
ΣFy= m•ay
ay=0, since the body is not accelerating
ΣFy = 0
Ng — Wf —W = 0
Ng = Wf + W
Ng = 490 + 810
Ng = 1300 N.
Also,
ΣFx= m•ax
ax=0, since the body is not accelerating
ΣFx = 0
Ff — Nw = 0
Ff = Nw
Now,
Let take moment about point A(ground), but note before we take moment, the forces must be perpendicular to the ladder.
applying condition of equilibrium of moment
Clockwise moment = anti-clockwise
WfCosθ•L2 + WCosθ•L1 = NwSinθ•L
(810Cos55)•3.9 + (490Cos55)•7 = (NwSin55)•14
1811.93 + 1967.37 = 11.47Nw
3779.3 = 11.47Nw
Then, Nw = 3779.3/11.47
Nw= 329.49N
Since Nw = Ff
Then Ff = 329.49N
So the required reaction exerted by the ground on the ladder are
Ng = 1300 N
Ff = 329.49 N
b. Now the firefighter is a distance of 9.4m from the horizontal and the ladder is about to slip
So we need to calculate the coefficient of static friction μs
Check attachment for new diagram,
So calculating for Nw again, since the firefighter have new position
Now the firefighter is at 9.4m from the ground. Therefore, L2 = 9.4m
applying condition of equilibrium of moment
Clockwise moment = anti-clockwise
WfCosθ•L2 + WCosθ•L1 = NwSinθ•L
(810Cos55)•9.4 + (490Cos55)•7 = (NwSin55)•14
4367.21 + 1967.37 = 11.47Nw
6334.58 = 11.47Nw
Then, Nw = 6334.58/11.47
Nw= 552.274N
Since Nw = Ff
Then Ff = 552.274N
Then, using frictional law
Ff = μs•Ng
The Ng doesn't change
Then, 552.274 = 1300μs
μs = 552.274/1300
μs = 0.42
c. Now, we want to know the maximum distance of the firefighter if the coefficient of static friction is reduce by half
Then, μs = 0.42/2
μs = 0.21
The assume the firefighter is at L2 from the horizontal.
Then, the frictional force is
Ff = μsNg
Ff = 0.21 × 1300
Ff = 273N
Then, Nw = Ff = 273N
Taking moment about point A
Clockwise moment = anti-clockwise
WfCosθ•L2 + WCosθ•L1 = NwSinθ•L
(810Cos55)•L2 + (490Cos55)•7 = (273Sin55)•14
464.6•L2 + 1967.37 = 3130.8
464.6•L2 = 3130.8—1967.37
464.6L2 = 1163.43
L2 = 1163.43/464.6
L2 = 2.5m
The maximum distance before the ladder begin to slip is 2.5m.
