# Analyzing Mass A's Direction of Motion

In summary, the problem asks for the time for A to slide 0.5 m relative to the ramp. The results will differ by a factor of 10.f

User has been reminded to show their best efforts when posting schoolwork-type questions
Homework Statement
The 10-kg block A rests on the 50-kg plate B .
Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide 0.5 m on the plate when the system is released from rest.
Relevant Equations
nothing special I am not very sure how I would be approach this.
Obviously it is stated in which direction it's going where we see that mass A goes to the right, but how do I determine this stuff analytically.

You have to show some effort in approaching this problem "I am not sure how to approach this" is not good enough.

How do we usually approach these kind of problems? What is one of the first things that we do?

• I did the analysis where they are in equilibrium then the tension force in A is way less than B, but how could this conclude that B goes down whilst A goes upP?

I did the analysis where they are in equilibrium then the tension force in A is way less than B, but how could this conclude that B goes down whilst A goes upP?
show work done

• You have to show some effort in approaching this problem "I am not sure how to approach this" is not good enough.

How do we usually approach these kind of problems? What is one of the first things that we do?
sorry oki,
equilibrium B with friction neglected.
##T_b = m_b*g/2 = 245.3N##
Eq. A with friction neglected:
##T_a = m_a*g/2 = 49.05N##
This is the place where there is actually no friction yet, just to correct miself.

Can I now think of this as to have A at rest, the rope needs to exert a 490.5N force, but at that point, mass B will already be moving such that it just goes upwards.

Why does it help to ignore friction?

What are the forces acting on A along the inclined plane?

What are the forces acting on B along the inclined plane?

• Why does it help to ignore friction?

What are the forces acting on A along the inclined plane?

What are the forces acting on B along the inclined plane?
don't care about those right now? or do we?
I mean they are let me calc real quick...

Why does it help to ignore friction?

What are the forces acting on A along the inclined plane?

What are the forces acting on B along the inclined plane?
well no, but I determine now what the direction of motion would be such that I can say that the friction opposes the motioin, isn't that a good way to do this.?

even if I knew the friction forces, that doesn't tell me nothing, because there's no direction I can associate with them right?

That's why I say Newton's laws for the static situation and see in which dir. it would move when there was no friction and then just oppose that motion. So to say Mass A goes up, thus friction towards the left.

By the way I did forget the normal force of A on B and of B on A in the analysis of equilibirum but is of no real consequence to the conclusion that 'd be drawn hand, at least I think ofc.

Define the direction positive down the incline, draw FBD for each object. Solve system of equation. If the forces you calculated have positive value they are directed down the incline, if the get negative they are directed up the incline.

The only thing you need to take care of is that the force of friction from A on B has reverse direction as the force of friction from B on A (Newtons third law). And that their accelerations will be equal in magnitude but opposite in direction (rope is stretched)

Last edited:
Define the direction positive down the incline, draw FBD for each object. Solve system of equation. If the forces you calculated have positive value they are directed down the incline, if the get negative they are directed up the incline.

The only thing you need to take care of is that the force of friction from A on B has reverse direction as the force of friction from B on A (Newtons third law). And that their accelerations will be equal in magnitude but opposite in direction (rope is stretched)
The thing is that this works very well for determining the static situation. However, in the dynamic situation the frictional forces are fixed and opposing motion. You therefore need to know the direction of motion to know the direction of the frictional forces. However, this can be inferred from the problem by, for example, looking at the no friction case (friction is not going to reverse the direction of acceleration). In this case though, the problem essentially gives you the direction of motion by stating the direction that the block A will move. This fixes the frictional forces and the unknowns are the tension and acceleration.

I will say that the problem statement is not entirely clear though. It is ambiguous if it asks for the time for A to slide 0.5 m relative to the ramp or relative to B. The results will differ by a factor of four.

• Define the direction positive down the incline, draw FBD for each object. Solve system of equation. If the forces you calculated have positive value they are directed down the incline, if the get negative they are directed up the incline.

The only thing you need to take care of is that the force of friction from A on B has reverse direction as the force of friction from B on A (Newtons third law). And that their accelerations will be equal in magnitude but opposite in direction (rope is stretched)
no but that is what is not the case on this one, because the forces themselves are calculated by the constraint equation F = ##mu * N##.
This means that I can get only two equations that useful from Newton's laws.
And then I can either say that the rope is masseless such that the accelerations are the same but in the opposite direction or use absolute dependent motion from the length of the rope and their position to get the third eq. that says that ##a_b = -a_a## which is not really a necessary step as it can be seen that the accelerations are equal and opposite.

This means that I need to determine the direction of the friction from the get go.

The thing is that this works very well for determining the static situation. However, in the dynamic situation the frictional forces are fixed and opposing motion. You therefore need to know the direction of motion to know the direction of the frictional forces. However, this can be inferred from the problem by, for example, looking at the no friction case (friction is not going to reverse the direction of acceleration). In this case though, the problem essentially gives you the direction of motion by stating the direction that the block A will move. This fixes the frictional forces and the unknowns are the tension and acceleration.

I will say that the problem statement is not entirely clear though. It is ambiguous if it asks for the time for A to slide 0.5 m relative to the ramp or relative to B. The results will differ by a factor of four.
sorry didn't see your comment when you posted it. ( I kind of said the same tjhing in the intial part.)

oh okay that is interesting. So the problem gave it, but i wanted to see how I could've determined it on my own. But euh- why is that difference by factor of 4?

But euh- why is that difference by factor of 4?
That is a good question. Can you figure it out? The argument is relatively simple and doesn’t really require any computation except ##2^2=4##.

There are only two options, A goes down and B up. A goes up and B down.
Figure out the directions of the friction forces by relative motion of the surfaces.
Solve each system of equations.
There is only one option which will result in a physical result

but i wanted to see how I could've determined it on my own
Generally you cannot from the force equations alone because the system might already be moving in the opposite direction to that of the acceleration. However, here you start from rest so the motion will be in the direction of acceleration and friction will oppose this acceleration. The direction of acceleration may be inferred from the frictionless case. If you do this and get a negative acceleration, then friction is strong enough for there to be no acceleration at all (ie, the assumption of the frictional forces taking their max value is violated).

There is only one option which will result in a physical result (one will give wrong direction of the rope-force)
This is not generally applicable. Consider the frictionless case. Regardless of the direction of motion the tension in the rope will be the same. Adding a tiny amount of friction to this will mean that the tension in the rope changes with the sign of the change depending on the direction of motion. However, as long as the friction is small this will not affect the sign of the tension.

• malawi_glenn
This is not generally applicable
Yes I realized that, that is why that part was removed in my post. Regarding the rope tension.

• Delta2
It is ambiguous if it asks for the time for A to slide 0.5 m relative to the ramp or relative to B. The results will differ by a factor of four.
Should that be a factor of ##\sqrt 2##?

Should that be a factor of ##\sqrt 2##?
Yes you are right, I blame the lack of morning coffee when I wrote that.

• SammyS
Surely the ##50 \rm{kg} ## must descend, or the system is static. How can it ever be the case that the ##10 \rm{kg}## block pulls the ##50 \rm{kg} ## block up the hill under its own weight?

EDIT:
Starting from rest

Last edited:
Surely the 50kg must descend, or the system is static. How can it ever be the case that the 10kg block pulls the 50kg block up the hill under its own weight?
Antifriction!

• erobz
Surely the ##50 \rm{kg} ## must descend, or the system is static. How can it ever be the case that the ##10 \rm{kg}## block pulls the ##50 \rm{kg} ## block up the hill under its own weight?
Nobody has said otherwise and as stated this is given in the problem.

Outside of the problem formulation however, it is perfectly possible (although not when starting from rest) for the 50 kg block to be moving upwards and the 10 kg block to be moving down. The acceleration on the other hand will accelerate the heavier block downwards.

• erobz
Nobody has said otherwise and as stated this is given in the problem.

Outside of the problem formulation however, it is perfectly possible (although not when starting from rest) for the 50 kg block to be moving upwards and the 10 kg block to be moving down. The acceleration on the other hand will accelerate the heavier block downwards.
I thought the OP was having trouble figuring out the direction that the friction forces act between the blocks ##A## and ##B## for this problem, and was searching for some intuitive approach.

I thought the OP was having trouble figuring out the direction that the friction forces act between the blocks ##A## and ##B## for this problem, and was searching for some intuitive approach.
He wanted to analytically prove it.

• erobz
This situation is a variant of the inclined Atwood machine with friction present. Because the pulley is ideal, the tension on the two string segments is the same. We can consider the FBD of the two-mass plus rope system in which the external forces acting on the system are
1. ##m_A~f\sin\theta~## acting on ##m_A## down the incline
2. ##m_B~f\sin\theta~## acting on ##m_B## down the incline
3. ##f_k=\mu_{\text{AB}}(m_A+m_B)g\cos\!\theta~## acting on ##m_B## parallel to the incline
This is a one-dimensional problem as far as the kinematics are concerned. We can straighten the string to a straight line and rotate the whole thing by ##\theta## to get the equivalent FBD shown below Here is why I drew the friction to the left, in the same direction as the tension on ##m_B##. Say friction and gravity can be turned on and off at will. Initially they are both off and the system, we are told, is at rest.
1. If we only turn on gravity, the system will accelerate to the right.
2. If we only turn on friction the system will remain at rest.
3. As mentioned above by @Orodruin in #20, if we turn on both friction and gravity, the system cannot possibly accelerate to the left. If it did, then statement 1 would be violated when we trun the friction back down to zero.
Note that the FBD above affords easy calculation of the magnitude of the common acceleration. Knowing that, one can easily sort out the internal forces from FBDs of the individual masses.

We can straighten the string to a straight line and rotate the whole thing by θ to get the equivalent FBD shown below There is also friction between the blocks. This of course represents a 3rd law pair but once you straighten the string they act in the same direction because one is turned 180 degrees.

• kuruman
Another way to get the initial direction is to consider the height of the system's centre-of-gravity (CoG).

When released from rest, the CoG must move downwards, converting gravitational potential energy to kinetic energy (and to heat).

This happens when the height of the heavier object (B) decreases and the height of the lighter object (A) increases.

(If the CoG moved upwards, this would violate the conservation of energy: gravitational potential energy, kinetic energy and heat would need to be created out of nowwhere.)

• Orodruin
There is also friction between the blocks. This of course represents a 3rd law pair but once you straighten the string they act in the same direction because one is turned 180 degrees.
Yes, of course, this makes much more sense. I assumed without thinking that the frictions would cancel like the tensions although it didn't sit well with me because of what happens when the incline is frictionless but there is friction between blocks. My excuse of the day is that my wife was pressuring me to do something for her so I posted in a hurry.

• SammyS
My excuse of the day is that my wife was pressuring me to do something for her so I posted in a hurry. • • 