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:
Explanation:
mass of the ball = 146 g = 146 / 1000 = 0.146 kg
initial speed of the ball = 40.6 m/s
final speed of the ball = - 45.1 m/s
time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s
impulse, Ft = change in momentum = mv - mu = m (v-u)
F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N
Answer:
The answer to your question is letter B.
Explanation:
To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.
Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.
The same number there were on the reactant side, before the reaction.
