Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
to provide a greater output force. In other words, we can lift a load without having to lift the full weight of the load. Prove using a mathematical relation that Class II levers provide mechanical advantage. You will need to use the rotational equilibrium condition. It will be helpful to draw a picture indicating the movement of the load. Also, to simplify we can assume the force is still perpendicular to by considering the case when the load just starts to move.
1 answer:
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
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Answer:
(i) See attached image for the drawing
(ii) net force given in component form: (20, 20)N with magnitude:
Explanation:
First try to write all forces in vector component form:
The force F1 acting at 45 degrees would have multiplication factors of on both axes, to take care of the sine and cosine projections. Therefore, the:
x-component of F1 is
y-component of F1 is
As far as force F2, it is given already in x and y components, then:
x-component of F2 = 8 N
y-component of F2 = -6 N (negative meaning pointing down the y-axis)
Force F3 has only component (upwards) in the y-direction
x-component of F3 = 0 N
y-component of F3 =14 N
The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:
x-component of R = 12 + 8 = 20 N
y-component of R = 12 + 14 - 6 = 20 N
Therefore, the acceleration that the mass receives due to this force is given in component form as:
x-component of acceleration: 20 N / 5 kg =
y-component of acceleration: 20 N / 5 kg =
Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:
x-component of the velocity is:
y-component of the velocity is:
1)
2) 13.9 s
Explanation:
1)
The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.
It can be calculated using the equation:
(1)
where
G is the gravitational constant
is the Earth's mass
r is the distance of the object from the Earth's center
The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is
where
is the Earth's radius
Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):
2)
The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula
where
L is the length of the pendulum
g is the acceleration due to gravity at the location of the pendulum
Note that the period of a pendulum does not depend on its mass.
For the pendulum in this problem, we have:
L = 3 m is its length
is the acceleration due to gravity (calculated in part 1)
Therefore, the period of the pendulum is: