Mechanics question help please

Question

There is a fixed wedge whose smooth sloping face is inclined at 30 degree to the horizontal. A particle P of mass 2 kg rests on the sloping face of the wedge. A light inextensible string, passes over a smooth pulley at the top of the wedge and connects P to a particle Q of mass 3 kg which hangs freely. The sloping part of the string is parallel to a line of greatest slope of the face of the wedge. A constant force of magnitude 25 N acting down a line of greatest slope, is applied to P. Assuming that P has not reached the bottom of the slope and Q has not reached the pulley , calculate :
(A) the tension in the string
(B) the work done by the 25 N force while the speed of the particles increases from 1 m/s to 3 m/s.

Answer 1

Hi Orawan!

 

I think I see the situation here.  The only phrase that throws me off a bit is "parallel to the line of greatest slope."  Since, at those points, we are talking about a smooth sloping face with a constant incline (30^o), I am going to assume that the string attached to P is parallel to the slope, and then Q is hanging vertically, and then work the problem that way.  If that assumption is mistaken, please forgive me and let me know.

 

I am also assuming, since no information is given about it -- and I cannot determine any from the problem -- that the slope is frictionless.

 

Ropes and strings in situations like this generally, in my opinion, make the problem seem trickier than it is.  You have one string going down at an angle, another hanging vertically... you seem to have to tie a lot of things together with multiple equations and geometry.

 

But the way I always approach these problems is with the realization that a string, however it can be wrapped around pulleys and whatnot, is basically a one-dimensional object.  And, like any one dimensional line, it allows for motion is two directions.  In this case we can describe those two directions as:

 

1.  Q falls and P moves up the slope (which we can call the "negative" direction)

2.  Q rises and P moves down the slope (which we can call the "positive" direction)

 

With this convention, we can define a system that includes both particles, and determine its acceleration under the influence of the forces pulling it in the "positive" and "negative" directions.  With this system, the tension of the string becomes an internal force, and thus does not play in Newton's 2nd law; but once we solve for the acceleration, we can redefine the system quickly to make tension external, and then solve for it.

 

Let's determine the forces acting to pull the system of the two blocks in our conventional directions, defined above:

 

a.  There is the force of gravity on Q, which pulls the system in the "negative" direction, as defined above, with a force of magnitude m_Qg = (3kg)(9.8 m/s²) = 29.4 N

b.  There is the force of gravity down the plane on P, which pulls in the "positive" direction with a magnitude of m_pgsin(θ) = (2 kg)(9.8 m/s²)(sin(30^o)) = 19.6 N

c.  The 25 N applied force pulling (again, I assume parallel to the plane) in the "positive" direction.

 

With these forces identified (in magnitude and direction), we can then write Newton's 2nd law for our one-dimensional system of two particles:

 

F_net = 25 N + 19.6 N - 29.4 N = m_sysa = (m_P+m_Q)a = (5kg)a

 

With this expression, the acceleration of the two particles can be calculated (and it will be in the positive direction).

 

Then, to solve for tension (T) in the string, we can redefine the system to be just the hanging particle Q.  For this, we can set up Newton's 2nd law again for this new system, where -- say -- "up" is positive and "down" is negative.  For just Particle Q, T pulls up and gravity pulls down with 29.4 N.  Thus:

 

F_netQ = T - m_Qg  = T - 29.4 N = m_Qa = (3 kg)(the acceleration calculated above)

 

So this is now one equation with one unknown, T, which you can solve for.

 

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To compute the work done by ONLY the 25 N force, I think it is easiest to compute the distance the particles move as they accelerate from 1 m/s to 3 m/s.  Because this is effectively one-dimensional motion, we can use one-dimensional kinematics on our two-block system -- solving for displacement d.  As with any kinematics problem, if we are solving for one kinematic quantity, we need to know at least three others.  In this case, we have:

 

To find:  d

know:  v_o, v, a

 

where v_o = 1 m/s  (initial velocity)

          v = 3 m/s  (final velocity)

          a = value calculated above (at which the two-particle system accelerates)

 

Then we pick out the kinematic equation relating these four quantities:

 

v² = v_o²+2ad

 

And solve for d.  Once you do that, assuming again that the force is acting parallel to the slope, in the direction of displacement,  the work done by the 25 N force is just:

 

W₂₅ = Fd = (25 N)(displacement solved for above)

 

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As always, let me know if you want to talk about any of this further, or check any answers.  I hope this helps!