Based on the free-body diagram, the net force acting on this firework is
2 answers:
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Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.
(b) The force the ground exerts on the front set of wheels is 0.239 MN.
<h3>Center mass of the airplane</h3>The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.
<h3>Some assumptions</h3>- The wheels under the wind do not pass through the center line.
- The position of the front wheel is constant and it is zero mark (origin).
- The rear wheels are at 21.7 m mark
Position of the center mass of the plane is calculated as follows;
Let the position of the center mass, Xcm = y
the center mass is 3 m in front of rear wheels, that is
21.7 - y = 3
y = 21.7 - 3
y = 18.7 m
Xcm = 18.7 m
<h3>Mass of the plane at the position of the rear wheels</h3>Let the mass of the plane at front wheels = M1
Let the mass of the plane at rear wheels = M2
There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;
M1 + M2 = 177,000
M1 = 177,000 - M2
M1 = 177,000 - 152,529.95
M1 = 24,470.05 kg
<h3>Force exerted by the ground on the front wheel</h3>W = mg
W = 24,470.05 x 9.8
W = 239,806.5 N = 0.239 MN
Learn more about center mass here: brainly.com/question/13499822