Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
Fall at equal acceleration with similar displacements.
Explanation:
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Objects under free fall with no air resistance, are falling under the sole influence of gravity. So, under that conditions, objects with different masses will fall with the same rate of acceleration
Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s