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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Complete question is;
Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.
Block 1 is removed without disturbing block 2.
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is mass of block 3
Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a
Making T the subject gives us;
T = (m2)g - (m2)a - - - (eq 2)
where;
g is acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is mass of block 2
To solve for the acceleration, we will just substitute (m3)a for T in eq 2.
Thus;
(m3)a = (m2)g - (m2)a
(m3)a + (m2)a = (m2)g
a(m3 + m2) = (m2)g
a = (m2)g/(m3 + m2)
