What is the most common measuring system in the world?
1 answer:
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Answer:
the force P required for impending motion is 132.3 N
the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
Explanation:
Given that:
mass of the cabinet m = 45 kg
coefficient of static friction μ = 0.30
A free flow body diagram illustrating what the question represents is attached in the file below;
The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.
Thus; considering the forces along the vertical axis ; we have :
The upward force and the downward force is :
where;
are the normal contact force at center point A and B respectively .
------- equation (1)
Considering the forces on the horizontal axis:
where ;
are the static friction at center point A and B respectively.
which can be written also as:
replacing our value from equation (1)
P = 0.30 ( 441)
P = 132.3 N
Thus; the force P required for impending motion is 132.3 N
b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm
the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at and it is shifted to
:
Then:
h = 0.8 m
Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.
(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