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12.Two masses are arranged as shown on a ramp that makes an angle of 36° with the horizontal. Mass m1 is twice mass m2
. What is the minimum coefficient of static friction required between mass m1 and the ramp to ensure equilibrium?

12Two masses are arranged as shown on a ramp that makes an angle of 36 with the horizontal Mass m1 is twice mass m2 What is the minimum coefficient of static fr class=

Respuesta :

Force  M2 exerts on tension T

T = M2 g

T = M1 g sin theta - M1 g u cos theta

Note that M1 g is exerted down the plane while the frictional force opposes this motion and will be directed in the opposite direction of M1 g

M2 g = M1 g sin theta - u M1  g cos theta

1 = 2 sin 36 - u 2 cos 36

.5 = sin 36 - u cos 36

u = (sin 36 - .5) / cos 36 = .108

Any friction less that this will result in sliding

Cricetus

"1.7456" would be the minimum coefficient of static friction required.

Given:

Angle,

  • [tex]\Theta = 30^{\circ}[/tex]

Mass,

  • [tex]m_2 =2 m_1[/tex]

To keep the system in equilibrium forces acting on mass "[tex]m_1[/tex]" must balance forces acting on "[tex]m_2[/tex]".

That's,

→ [tex]Forces \ on \ m_1 \ downward \ along \ ramp = Forces \ on \ m_2 \ downwards[/tex]

or,

→ [tex]m_1 g Sin \Theta + \mu_s m_1 g Cos \Theta =m_2 g[/tex]

Thus,

→ [tex]\mu_s = \frac{m_2 g - m_1 g Sin \Theta}{m_1 g Cos \Theta}[/tex]

       [tex]= \frac{m_2-m_1 Sin \ 36^{\circ}}{m_1 Cos \Theta}[/tex]

hence,

→ [tex]\mu_s = \frac{2 m_1 - m_1 Sin \ 36^{\circ}}{m_1 Cos \ 36^{\circ}}[/tex]

       [tex]= 1.7456[/tex]

Thus the above answer is right.

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