Which of the following can be calculated when given the radius of a wheel and the radius of an axle?
2 answers:
Answer:
C. the mechanical advantage of the wheel and axle
Explanation:
the ratio of the radius of a wheel to the ratio of an axle gives the mechanical advantage for that wheel and axle system
i.e Mechanical advantage =
for example if the radius of a wheel is 70m and the radius of an axle is 35m, the mechanical advantage for that machine is
M.A = 70/35 = 2
This implies that the outpu force is 2 times that exerted by the input force
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The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force 
which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
, 
. Re-arranging, we can solve this equation to find 
, the coefficient of dynamic friction:
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
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
