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3 years ago

Find the force needed to lift a 3000 N weight using a machine with a mechanical advantage of 15.

2 answers:

8 0

IF the machine is 100% efficient, then you only need to push at the input with a force of 200 Newtons. BUT ... you have to keep pushing the 200N 15 times as far as you want to lift the big weight.

slega [8]3 years ago

7 0

Answer: 200N

Explanation: Mechanical advantage is the ratio of how the machine transforms the force that you apply in it.

Ma = Fb/Fa

Where Fb is the force that the machine does and Fa is the force that you do to the machine.

If the mechanical advantage is 15, this means that the force that you apply at the machine is multiplied by 15.

If we want to lift an object that weighs 3000N, the minimal force that we need to apply is also 3000N. so we need to solve the equation:

Fb = 15*Fa = 3000N

Fa = 3000N/15 = 200N

So the force that we need to apply is 200N

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Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

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d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

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