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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
of 30' with the horizontal. If the coefficient of friction between the wheels of the carriage and the grass is
0.14, calculate the weight, normal force, frictional force and the acceleration of the carriage.

Respuesta :

Weight of the carriage [tex]=(m+M)g =142.1\ N[/tex]

Normal force [tex]=Fsin(\theta) + W = 197.1\ N[/tex]

Frictional force [tex]=\mu N=27.59\ N[/tex]

Acceleration [tex]=4.66\ m\ s^{-2}[/tex]

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) [tex]=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)[/tex]

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with [tex]F_x[/tex], force of [tex]110\ N[/tex] acting vertically downward.Both are downward and Normal is upward so Normal force [tex]=Summation\ of\ both\ forces[/tex]

  • Normal force (N) [tex]= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N[/tex]
  • Frictional force (f) [tex]=\mu N=0.14\times 197.1 =27.59\ N[/tex]

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that [tex]F_y=Horizontal[/tex] and [tex]F_x=Vertical[/tex] component of forces.

So Fnet = Fy(Horizontal) - f(friction) [tex]= m\times a[/tex]

  • Acceleration (a) =[tex]\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }[/tex]

So we have the weight of the carriage, normal force,frictional force and acceleration.

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