Relative Acceleration Problem

**Dorothy Weglend** · Oct8-06, 10:41 AM

This is problem 5-61 in Serway and Jewett, 4th Ed. (See attached figure).

The problem is to find the force F which will keep the blocks stationary relative to the cart. All surfaces, wheels and pulley are frictionless.

I get F=(m1+m2+M)(g*m2/(m1+m2)

The answer in the back of the book is

F=(m1+m2+M)(g*m2/m1)

Is this right? Here is my work (T is tension in the string):

T - m1*a = 0
m2*g - T = m2*a

Solving for a: a = g*m2/(m1+m2)

The solution in the book must have come from these two equations:

T - m1*a = 0
m2*g - T = 0

which are from the point of view of an observer outside of the cart, I guess.

I realize that if my solution works, the books will also, since it gives a larger F, but will the smaller acceleration not be enough?

Perhaps I don't understand inertial frames very well yet.

Thanks,
Dorothy

**Dorothy Weglend** · Oct8-06, 01:27 PM

I have thought more about this problem, and it seems harder than I thought...

I see that I have neglected the acceleration of the cart, and this isn't really an inertial frame problem. I wonder if this problem is even solvable...

Considering the acceleration of the cart, for m1 we have

T = m1(a-A)

And for the vertical force on m2

m2g - T = m2a

But it seems there must be an additional component of tension in the horizontal direction, because the string is being pushed on by m2 with force F (A(m1+m2+M).

So this must increase the tension on T, so the equation for m1 becomes:

T + A(m1+m2+M) = m1(a-A)

we want a-A = 0, so

T + A(m1+m2+M) = 0
m2g-T = m2a

Adding I get m2g + A(m1+m2+M) = m2a
or

m2g + F = m2a

so F = m2(a-g).

Urgh... This doesn't seem right.

Please somebody, help me out :-)

Thanks,
Dorothy

**Dorothy Weglend** · Oct9-06, 01:18 AM

Ok, I get it, never mind.... The cart accelerates, which gives an acceleration to m1. But since it accelerates enough that the blocks stay in the same place, there is no acceleration for the hanging mass, m2.

So the solution as given in the book seems correct...

Well, I'll check back to see if there are any illuminating comments from anyone.

Thanks,
Dorothy

**Dorothy Weglend** · Oct9-06, 02:36 AM

In my next problem, we are asked to find the tension on the string in the general case for the system shown in the diagram. Everything is held motionless, and then, at the instant after the release of the system, we want to find the tension, etc.

I have solved this problem, and even managed to get the same answer given in the back. But it has caused me to question my first answer.

To solve it, in addition to the masses, I consider the motion of the cart at the pulley. So for m1

T= m1(a-A)

where a is the acceleration of the m1 from the string, and A is the acceleration of the cart.

At the pulley, T = MA.

So we have:

m1(a-A) = MA

But in the original problem, the cart accelerated enough so that the blocks remain motionless relative to the cart, so a-A = 0, which means...

0 = MA

This would seem to lead to a contradiction: If the only force on the cart that would cause no motion in the blocks is zero, well, the blocks will accelerate due to gravity.

I hope somebody can explain this to me...

Thanks thanks thanks.
Dorothy

**Ng Xin Zhao** · Dec30-06, 04:01 AM

In the general case, the whole system should not be accelerating. And m2 have acceleration,a downwards. Then we have
m2*g-T=m2*a
m1 will have the same acceleration,a and we have
T-m1*a=0
solving for T, we have T=m1*m2*g/(m1+m2)

**PhanthomJay** · Jan2-07, 08:55 AM

Originally Posted by Dorothy Weglend

In my next problem, we are asked to find the tension on the string in the general case for the system shown in the diagram. Everything is held motionless, and then, at the instant after the release of the system, we want to find the tension, etc.

I have solved this problem, and even managed to get the same answer given in the back. But it has caused me to question my first answer.

To solve it, in addition to the masses, I consider the motion of the cart at the pulley. So for m1

T= m1(a-A)

where a is the acceleration of the m1 from the string, and A is the acceleration of the cart.

At the pulley, T = MA.

So we have:

m1(a-A) = MA

But in the original problem, the cart accelerated enough so that the blocks remain motionless relative to the cart, so a-A = 0, which means...

0 = MA

This would seem to lead to a contradiction: If the only force on the cart that would cause no motion in the blocks is zero, well, the blocks will accelerate due to gravity.

I hope somebody can explain this to me...

Thanks thanks thanks.
Dorothy

In part 1, with the top block motionless with respect to the cart, the FBD of block 1 is
T=m1(a1), where a1 is the acceleration of the block with respect to the ground. For the hanging block 2, the FBD is
m2(g) - T = m2(a2), where a2 is the downward acceleration of the block, and must be equal in magnitude to the horizontal acceleration of the block m1 with respect to the cart, since the cord is assumed massless and inextensible. Thus, a2 = 0, solve T =m2(g), a1 = (m2/m1)g, and by taking a FBD of the system, F = (m1 +m2 + M)(m2/m1)g. Note that it's OK in this special case to take a FBD of the system, since the accelertaions of all blocks and cart horizontally are the same with respect to the ground. In general, however, this is NOT the case, and a FBD of the system will not work. You must instead take a FBD of the cart and hanging block, a bit tricky, but, thus
F -T = (M + m2)a_cart, which , for this special case, since a_cart is a1, and T = m2(g), yields
F = T + (M + m2)a1
F = m2(g) + (M +m2)(m2/m1)g, which will yield the same result for F.

For the general case, you must take a FBD of the top block , T =m1(a1), where a1 is the acceleration of the block with respect to the ground. Then an FBD of the hanging block, m2(g) - T = m2(a2), where a2 , the acceleration of the hanging block, is a2 = a_cart - a1, and finally an FBD of the cart and hanging block, F - T = (M + m2)a_cart. Solve these equations in terms of F, the applied force. You cannot take an FBD of the system for this general case. Note also that for certain values of F, the cart moves backwards, and when F = T, the cart is stationary, and you get 'Ng's" result.

**Dorothy Weglend** · Jan2-07, 11:33 AM

Oh wow, I left those messages 2 or 3 months ago. Thanks, Zhao and Phantom, for your help.

Dorothy

**vij** · Jan2-07, 12:42 PM

Originally Posted by Dorothy Weglend

This is problem 5-61 in Serway and Jewett, 4th Ed. (See attached figure).

The problem is to find the force F which will keep the blocks stationary relative to the cart. All surfaces, wheels and pulley are frictionless.

I get F=(m1+m2+M)(g*m2/(m1+m2)

The answer in the back of the book is

F=(m1+m2+M)(g*m2/m1)

Is this right? Here is my work (T is tension in the string):

T - m1*a = 0
m2*g - T = m2*a

Solving for a: a = g*m2/(m1+m2)

The solution in the book must have come from these two equations:

T - m1*a = 0
m2*g - T = 0

which are from the point of view of an observer outside of the cart, I guess.

I realize that if my solution works, the books will also, since it gives a larger F, but will the smaller acceleration not be enough?

Perhaps I don't understand inertial frames very well yet.

Thanks,
Dorothy

Your answer is correct. The acceleration produced in m1 and m2 by the driving force m2*g should be equal to the acceleration produced in the entire system by the driving force F. Therefore, you have, m2*g/(m1+m2) = F/(m1+m2+M) from which F = m2*g(m1+m2+M)/(m1+m2)

**PhanthomJay** · Jan2-07, 10:36 PM

Originally Posted by vij

Your answer is correct. The acceleration produced in m1 and m2 by the driving force m2*g should be equal to the acceleration produced in the entire system by the driving force F. Therefore, you have, m2*g/(m1+m2) = F/(m1+m2+M) from which F = m2*g(m1+m2+M)/(m1+m2)

No, that is not correct. I didn't realize this was so old a post, but in any case I was responding to the general case in part b. I believe it has been established for part a of this problem, where the blocks are at rest with respect to the cart, that the book answer is correct. To repeat part 1, the FBD of hanging block m2 leads to T=m2(g), then the FBD of the upper block m1 must lead to T = m1(a), where a is the acc. of that block with respect to the ground, and is equal also to the cart's acceleration. Thus m2(g) = m1(a) , solve a = (m2/m1)(g), then an FBD of the system, allowed since all the horizonatl accelerations of the blocks and cart are the same, yields the book result, the correct answer.

**vij** · Jan3-07, 12:03 AM

Originally Posted by Dorothy Weglend

I have thought more about this problem, and it seems harder than I thought...

I see that I have neglected the acceleration of the cart, and this isn't really an inertial frame problem. I wonder if this problem is even solvable...

Considering the acceleration of the cart, for m1 we have

T = m1(a-A)

And for the vertical force on m2

m2g - T = m2a

But it seems there must be an additional component of tension in the horizontal direction, because the string is being pushed on by m2 with force F (A(m1+m2+M).

So this must increase the tension on T, so the equation for m1 becomes:

T + A(m1+m2+M) = m1(a-A)

we want a-A = 0, so

T + A(m1+m2+M) = 0
m2g-T = m2a

Adding I get m2g + A(m1+m2+M) = m2a
or

m2g + F = m2a

so F = m2(a-g).

Urgh... This doesn't seem right.

Please somebody, help me out :-)

Thanks,
Dorothy

The correct answer indeed is F = m2g(m1+m2+M)/m1.
The answer I gave in a quick (in fact, too quick!) response was incorrect with m1+m2 in the denominator, in place of m1.
The masses m1 and m2 are at rest relative to the cart and hence their acceleration is zero. So we have the following two equations for the tension:
(1) T = m2*g
(2) T = m1*a where 'a' is the acceleration of the entire system consisting of the cart(of mass M) and the masses m1 & m2 given by
a = F/(m1+m2+M)
Therefore, from equations (1) and (2), m2*g = m1*F/(m1+m2+M) from which
F= m2*g(m1+m2+M)/m1.

**PhanthomJay** · Jan3-07, 12:23 AM

Originally Posted by vij

T = m1*a where 'a' is the acceleration of the entire system ...{snip}

It is true in this special case that 'a' is the acceleration of the entire system. However, in general, 'a' in this equation is the acceleration of the upper block with respect to the ground, which, in general, is unequal to the carts acceleration with respect to the ground, and in which general case, you introduce complexities when talking about a 'system' acceleration.

**Ng Xin Zhao** · Jan3-07, 05:42 AM

I am sorry but this is getting more and more confusing and the equations we write cannot be seen clearly (I don't try to read them anyway). So I am out of here thanks and call me Xin Zhao please.

**vij** · Jan3-07, 09:38 AM

I was referring to the special case of finding the force F which will keep the blocks stationary relative to the cart. In the general case, the situation is entirely different as you pointed out.

**atat1tata** · Mar26-08, 05:53 AM

I am fighting with that problem, either. My question is: if there is no friction at all and I push the pulley the mass above will experience only the tension of the string, right?
So:

(1) m1*a = T
(2) M*a = F - N
(3) m2*a = N
(4) m2*(ay)= 0 = m2g - T

(combining (2) and (3))
F = (M+m2)*a
(combining (1) and (4))
a = m2g/m1
And:
F = (M+m2)*g*m2/m1

I can see what is wrong here, since reasonably the force is proportional to the sum of the masses, but this is the outcome of my FBDs. What is my mistake?

**Doc Al** · Mar26-08, 06:14 AM

Originally Posted by atat1tata

(1) m1*a = T
(2) M*a = F - N
(3) m2*a = N
(4) m2*(ay)= 0 = m2g - T

Your error is with equation (2), where you assume that the only horizontal forces on M are F and N. (Don't forget that the pulley is attached to M.)

**atat1tata** · Mar26-08, 08:03 AM

Thanks a lot! So it is actually

(2) M*a = F - N - T

And this is so because the string (I believe) pulls on the pulley, and thus on M, downwards and leftwards, each with the magnitude of T.

(1+2+3) F = (m1+m2+M)*a

And everything seems correct. Thank you for helping me. I thought about this problem for hours.