What does mechanical advantage describe regarding simple machines?
2 answers:
Answer:
They multiply the input force.
Explanation:
Mechanical Advantage is the multiplication of the input force to give an increased output by a <em>simple machine </em>due to the difference in the length of load arm and the effort arm.
It is based upon the law of conservation of moments as to force applied at the farther point from a hinge or a pivot needs to be balanced by a greater force counteracting at a shorter distance from the pivot.
<u>Lever is a simple machine which consists of a bar pivoted or supported at a point (called fulcrum) about which it can rotate.</u>
..............................................(1)
where:
force applied on the end of effort arm
force due to load on the end of load arm
L = length of load arm
E = length of effort arm
Equation (1) is known as the law of levers, it was proven by <em>Archimedes </em>using geometric reasoning.
You might be interested in
Answer:
The weight of the object X is approximately 3.262 N (Acting downwards)
The weight of the object Y is approximately 8.733 N (Acting downwards)
Explanation:
The question can be answered based on the principle of equilibrium of forces
The given parameters are;
The weight of Z = 12 N (Acting downwards)
The weight of the pulleys = Negligible
From the diagram;
The tension in the in the string attached to object Z = The weight of object Z = 12 N
The tension in the in the string attached to object X = The weight of the object X
The tension in the in the string attached to object Y = The weight of the object Y
Given that the forces are in equilibrium, we have;
The sum of vertical forces acting at a point, = 0
Therefore;
Where;
= The weight of object Z = 12 N
= 12 N
= The vertical component of tension, T₂ = T₂ × sin(24°)
∴ = T₂ × sin(156°)
Similarly;
= T₃ × sin(50°)
From , and
= 12 N, we have;
12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)
Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0
Therefore at point B, we have;
T₁ₓ + T₂ₓ + T₃ₓ = 0
The tension force, T₁, only has a vertical component, therefore;
∴ T₁ₓ = 0
∴ T₂ₓ + T₃ₓ = 0
T₂ₓ = -T₃ₓ
T₂ₓ = T₂ × cos(156°)
T₃ₓ = T₃ × cos(50°)
From T₂ₓ = -T₃ₓ, we have;
T₂ × cos(156°) = - T₃ × cos(50°)...(2)
Making T₃ the subject of equation (1) and (2) gives;
Making T₃ the subject of equation in equation (1), we get;
12 = -(T₂ × sin(156°) + T₃ × sin(50°))
∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))
Making T₃ the subject of equation in equation (2), we get;
T₂ × cos(156°) = - T₃ × cos(50°)
∴ T₃ = T₂ × cos(156°)/(-cos(50°))
Equating both values of T₃ gives;
(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))
-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))
∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283
∴ T₂ ≈ -8.02 N
From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;
T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717
∴ T₃ ≈ -11.4 N
The weight of the object X = -T₂ × sin(156°)
∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N
The weight of the object X ≈ 3.262 N (Acting downwards)
The weight of the object Y = -(T₃ × sin(50°))
∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N
The weight of the object Y ≈ 8.733 N (Acting downwards)
