A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c
2 answers:
Answer:
d=1.18m
Explanation:
A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the coefficient of static friction between the floor and the ladder is 0.28. what distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?
frictional force =uFn
u=coefficient of static friction
Fn=normal force
assuming forces in the y direction, we ave
EFy=0
Efy=Fn-Fg-Fp
Fn=Fg+Fp
Fn=120N+600N
Fn=720N
we recall tat Frctional force=uFn
0.28*720N
201.6N
summing te forces in te x direction
EFx=Fr-fWall
Fr=frictional force
Fwall=force in the wall
Fr=Fwall
Fwall=201.6N
the angle te ladder makes wit round will be
cos=3/5
=53.13
Next, we can turn to calculating the net torque
Choose the pivot point at the bottom of the ladder
(this choice eliminates FN and Fr).
rgrav =1/2*5=2.5
rwall=5m
rman=d
twall== r( wall)(Fwall)sinalpa
5*201.6sin53.13
twall=+806.39Nm
τ grav = r( grav )(Fgrav )sin53.13
τ grav=2.5*120*)sin53.13
-239.99nm
τ person = r( person )(Fperson )sin53.13
d*600Nsn53.13
-479.99d
∑τ = τ wall + τ grav + τ person
= 0
summation of te torque
te negative sin is for troques in te clockwise direction
+806.39Nm-239.99nm-479.99d=0
566.4=479.99
d=1.18m
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a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.