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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:
Explanation:
<u>Circular Motion </u>
Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:
Where is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:
We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call the angular speed of the loaded reel and
that from the empty reel. We have
If is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by
The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.
If is the radius of the reel when it's empty, the angular speed for the empty reel is computed by
The full reel goes from to
in 6480 seconds, so we can compute the angular acceleration:
The empty reel goes from to
in 6480 seconds, so we can compute the angular acceleration:
So the equations for both reels are
They will be the same when
Solving for t
The common angular speed is
They both result in the same, as expected