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
velocity of the physics instructor with respect to bus
acceleration of the bus is given as
acceleration of instructor with respect to bus is given as
now the maximum distance that instructor will move with respect to bus is given as
so the position of the instructor with respect to door is exceed by
so it will be moved maximum by 3 m distance
Answer:
Explanation:
We have the following information,
We apply the equation for capacitor charging the voltage across it,
Replacing values,
<span>The weightlifter does no work. Although he has exerted force, work is the product of force over distance. Since he has not moved the wall he has done no work.</span>
