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There are two boxes with same mass couple together by a massless rope. One of the masses is on a ramp raised to an angle of 42 degrees, the other mass is hanging off to the right as shown in the figure. In between the two masses, the rope sun over a frictionless, massless pulley. If the mass on the ramp starts the problem at the top of the inclined plane, and the system is given a push, so that it is moving with an initial velocity pointing down the ramp. Ignore wind resistance for this problem.

a) If the coefficient of friction is 0.30, what is the magnitude and direction of the acceleration of the system? Use the coordinate system given in the figure.

b) Explain using a few sentences to answer the following: If the ramp is very long will the system eventually come to a stop?

c)What type of friction is present in this problem, Static, Kinetic, or Rolling?

d)What is the angle required for the the acceleration of the system to equal zero? Explain you reasoning with a few sentences.

Expert Solution

Step 1

The mass on the inclined ramp has a gravitational force acting on it vertically downwards.

This gravitational force can be resolved into two components.

One component parallel to the ramp along the -X direction, $m g \sin θ$

Other component perpendicular to the ramp along -Y direction, $m g \cos θ$

As the mass moves, force of friction opposes the motion of the mass, and thus acts opposite to the motion, along +X.

The tension in the string is T, and it is directed opposite to the motion of the mass m on the ramp.

The tension in the string for the other mass m is also T and is directed upwards in the string.

Step 2

The net force acting on the mass m on the ramp is along the -X direction and is equal to

$m a = m g \sin θ - T - F_{s} F_{s} i s t h e f o r c e d u e t o f r i c t i o n a n d i s g i v e n a s F_{s} = μ F_{N} μ i s t h e c o e f f i c i e n t o f f r i c t i o n a n d F_{N} i s t h e n o r m a l f o r c e a c t i n g o n t h e m a s s m T h e n o r m l f o r c e a c t i n g o n t h e m a s s m i s e q u a l t o a n d o p p o s i t e l y d i r e c t e d t o t h e p e r p e n d i c u l a r c o m p o n e n t o f t h e g r a v i t a t i o n a l f o r c e F_{N} = m g \cos θ F_{s} = 0.3 m g \cos θ m a = m g \sin θ - T - 0.3 m g \cos θ$

The net force acting on the mass m on the other side is in the upward direction along the tension and is equal to

$m a = T - m g$

Adding the above two equations,

$2 m a = m g \sin θ - T - 0.3 m g \cos θ + T - m g 2 m a = m g [\sin θ - 0.3 \cos θ - 1] 2 a = g [\sin 42 - 0.3 \cos 42 - 1] 2 a = g [0.6691 - 0.2229 - 1] 2 a = g [- 0.5538] 2 a = 9.8 [- 0.5538] a = \frac{- 5.42724}{2} a = - 2.71362$

The magnitude of the acceleration is 2.71362 m/s2 and it is along the -X direction

Step 3

If the ramp is long enough, at a certain point, the mass on the other side will reach the end of the pulley.

After this point, the mass will go no further and as a result the other mass would also stop moving.

So, the system will eventually come to a stop

Step by step

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