Answer:
Torque=13798.4 N.m
Explanation:
Given data
Mass of beam m₁=500 kg
Mass of the person m₂=70 kg
length of steel r₁=4.40m
center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m
To find
Torque
Solution
Torque due to beam own weight

Torque due to person

Now for total torque

Answer:
Work done against gravity will be
W = Mgℓ
Explanation:
Work done to raise the mass from ground to given height is against gravity
So here work done is given by the formula

here we know that

it is the force due to gravity which is also known as weight
so here distance moved by the object is given as
d = ℓ
now work done is given as
W = Mg ℓ
Angular velocity of the rotating tires can be calculated using the formula,
v=ω×r
Here, v is the velocity of the tires = 32 m/s
r is the radius of the tires= 0.42 m
ω is the angular velocity
Substituting the values we get,
32=ω×0.42
ω= 32/0.42 = 76.19 rad/s
= 76.19×
revolution per min
=728 rpm
Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.