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Alecsey [184]
3 years ago
10

Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1

x and F2x represent the components, of the corresponding forces. Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, of the corresponding forces. F1x=−F2x F1x=F2x m1=m2 m1≪m2
Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

a) m₁ = m₂  F₁ₓ = F₂ₓ

b) m₁ << m₂   F₂ₓ =0

Explanation:

This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law

     ∑ F = m a

for acceleration to be zero implies that the net force is zero.

we must write the expression for the center of mass

        x_{cm} = 1 / M (m₁ x₁ + m₂ x₂)

now let's use the derivatives

      a_{cm} = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)

where M is the total mass M = m₁ + m₂

     so that the acceleration of the center of mass is zero

               0 = 1 / M (m₁ a₁ + m₂a₂)

               m₁ a₁ = - m₂ a₂

In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore

   F₁ₓ = -F₂ₓ

b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂

      acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)

so that the acceleration of the center of mass is zero

               0 = 1 / M (m1 a1 + m2a2)

               m1 a1 = - m 2 a2

with the initial condition, we can despise m₁, therefore

                0 = m₂a₂

 if we use Newton's second law

              F₂ = 0

       

I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
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Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

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i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

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Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

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t = 300 / 22500

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Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

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F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

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F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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