Mechanics question please help

Question

(A) Two particles P and Q of masses 10 kg and 5 kg respectively are attached to the ends of a light in-extensible string. The string passes a fixed smooth peg. ( It's something like two particles hanging from a pulley). Initially particle Q is pulled vertically downwards with a speed of 1 m/s. Find the time elapse when the system is instantaneously at rest.

 

 

(B) Particle P of mass 2 kg is on a rough horizontal plane and particle Q of mass 1 kg on a rough plane inclined at 30 degree with the vertical. The particles are attached to the ends of a light in-extensible string which passes over a fixed smooth pulley C. Particle P, particle Q, the string and the pulley are in vertical plane. The particles are released from rest with the string taut and Q moves down the inclined plane. If the coefficient of friction of both the horizontal and the inclined plane is z, find the exact value of the acceleration of particle Q in terms of z.

Answer 1

Hi Orawan!

 

Whenever two (or more) objects are attached to a rope and slung over a pulley or a peg in any configuration (both hanging; one hanging and one on an incline; one hanging and one on a horizontal surface; to name a few examples), it is a configuration that seems more involved than it is.  The two particles may be moving in different directions, and you have to set up force equations for each one and connect them, and--

 

But a key thing about ropes is that, whatever weird way they may be wrapped, they are (for our purposes) one-dimensional objects.  Masses connected by a taut, inextensible rope can only move in two directions, just like objects on a one-dimensional line or axis.  So, whichever way the rope is wrapped, we can still consider this one-dimensional motion, and use whatever tools we have for that.

 

In addition, as long as the rope is taut and inextensible, so that the two particles must always be moving the same way with the same acceleration, we can consider the two particles (and the rope between them) as a system.  This has the advantage of making the tension of the rope an internal force, so we do not have to worry about it in Newton's laws (as long as we do not have to solve for it).

 

So, a way to picture this situation in the (A) case, where the two particles are slung over a peg, and each can move vertically, is to imagine taking those two particles and the rope between them and, keeping the rope taut, laying them along a flat, frictionless surface.  In this picture, any force that pulls Q down (and thus P up) acts to the right, which we can call positive, and any force that pulls P down (and thus Q up) acts to the left, which we can call negative.  This is conceptually making the system's one-dimensional motion manifest in a way we may be more used to seeing for one-dimensional motion.

 

With this in mind, let's look at Question (A):

 

Because we have one-dimensional motion, we can actually treat this system with one-dimensional kinematics.  As with any kinematics problem, we are trying to solve for one kinematic quantity (time, t).  If we are to use one of the standard set of kinematic equations -- each of which involves four of the five kinematic quantities -- then we need to know at least three other of the kinematic quantities (initial velocity, v_o; final velocity, v; acceleration, a; displacement, d).  For this case, let's figure out what we know:

 

to find: t

know:  v_o, v

 

We are told the system is given an initial velocity that involves Q moving vertically downward (which we called "to the right," or positive, in our conceptual laying out of the system on a horizontal frictionless surface).  So, v_o = (+) 1 m/s.  We are looking at an interval at the end of which the system is (momentarily) at rest, so the final velocity v = 0.

 

But we need to know one more kinematic quantity to solve for t.  Our choices are displacement, about which we are told nothing, and acceleration.  Because we know about the weights of the particles, we can get to the acceleration through Newton's second law, so let's concentrate on that.

 

First, what external force(s) is or are acting on our one-dimensional system?  With no friction from the peg, the only external force that sets the system in motion is gravity acting on each particle.  The force of gravity acting on particle P pulls it down, which is "left" in our conceptual horizontal picture (and thus negative).  The force of gravity acting on particle Q pulls it down, which is "right" in our conceptual horizontal picture (and thus positive).

 

Now we can set up Newton's second law along the one dimension of rope motion:

 

F_net = -F_gonP + F_gonQ = -(10 kg)(9.8 m/s²) + (5 kg)(9.8 m/s²) = -98 N + 49 N = - 49 N = m_sysa

 

where

 

m_sys = mass of the system = mass of both particles together = 15 kg

 

Now we can solve for the acceleration of the system:

 

- 49 N = (15 kg)a --> a = -3.27 m/s²

 

So now we know the acceleration of the system, which is the third kinematic quantity we need above to solve for the time, t, for the system to come momentarily to rest.  We have:

 

to find: t

know: v_o, v, a

 

v_o = (+) 1 m/s

v = 0 m/s  (at rest)

a = -3.27 m/s²

 

Now we just choose the kinematic equation that involves these four quantities.  It is:

 

v = v_o + at

 

And solve for t.  If you would like to check an answer, just let me know!

 

Now, for Question (B)

 

Even with the odd configuration, the concept of dealing with this as one dimensional motion still holds.  We can imagine this system on a horizontal, frictionless plane, where any force that acts to pull Q down its incline (and thus pull P in the corresponding direction along the rough horizontal surface, is "to the right," or positive;  and any force that acts to hold P or Q back from moving that way, or pulls them the opposite way, acts "to the left," or negative.

 

If I have this pictured correctly (without a diagram, I cannot be 100% certain), Particle Q (which I imagine as being to the right of the Particle P, is on a incline that is 60^o below the horizontal (hence 30^o from the vertical).  As a result, the force of gravity pulling it down the plane is F_dp = mgsin(θ) = (1 kg)(9.8 m/s²)(sin(60^o)) = 8.49 N.  I can derive that expression for force down an inclined plane, if needed.  This force acts in our positive direction.

 

There is then a force of friction on each block, which acts in our "negative" direction.  In each case, if the blocks are sliding, this must be kinetic friction, so that:

 

F_k = μ_kN

 

where

 

μ_k = coefficient of (kinetic) friction = z (for each particle)

N = normal force on the surface on the particle

 

So, to make an expression for the friction force, we have to calculate the normal force N on each particle.

 

On Particle P, the normal force is in the vertical direction.  Since P is not accelerating vertically, then another force must be cancelling out the normal force exactly.  That force is the force of gravity, so the magnitude of the normal force equals the weight of the particle.  N_P = (2 kg)(9.8 m/s²) = 19.6 N

 

For Particle Q, the normal force is perpendicular to the surface of the incline (by definition; it must always be perpendicular to the surface exerting it).  Since Q is not accelerating perpendicular to the incline, that normal force must be balanced out by another force in the opposite direction.  This is the component of the the force of gravity on Q that is perpendicular to the incline.  I will not derive it here, but that component of the gravitational force perpendicular to an incline is given by F_pp = mgcosθ) = (1 kg)(9.8 m/s²)(sin(60^o)) = 4.9 N.  Thus, the magnitude of the normal force on Q, N_Q = 4.9 N

 

With these values and z, we can calculate the forces of kinetic friction on each particle.  Both of these forces point to the left in our conceptual horizontal picture, and are thus negative.

 

Now, we can set up Newton's second law for our conceptual one-dimensional system:

 

F_net = F_dp-F_kP-F_kQ = 8.49 N - z(19.6 N) - z(4.9 N) = 8.49 - z(24.5) = m_sysa

 

Again, m_sys is just the mass of both particles together, which is 3 kg here.  So

 

24.5z = (3 kg)a -->  a = 8.17z

 

I hope this helps!  If you have any questions, just let me know, or if I misunderstood the problem.