Answer:
a) Impulse |J|= 219.4 kgm/s
b) Force F = 2672 N
Explanation:
Given
Height of fall h = 0.50 m
Mass M = 70 kg
Period of collision t = 0.082 s
Solution
The final velocity of the person v is zero since the person will come to rest.
The initial velocity of the person can be calculated by using the "law of conservation of energy".
Initial Kinetic energy = Final potential energy
a) Impulse
J = final momentum - initial momentum
Magnitude of impulse
b) Force
Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α = =
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²
Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m
AC = distance of person on the ladder from the bottom end = x
W = weight of the ladder = 240 N
= weight of the person = 710 N
F = force by the wall on the ladder
N = normal force by ground on the ladder = ?
Using equilibrium of force along the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f =static frictional force on the ladder
Static frictional force is given as
f = μ N
f = (0.55) (950)
f = 522.5 N
Force equation along the horizontal direction is given as
F = f
F = 522.5 N
using equilibrium of torque about point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
