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Total Impulse of/on a block pushed along a surface for 10s

  1. Oct 7, 2015 #1
    • This homework problem was originally posted in another forum so there is no template
    Suppose a block of 5kg is stationary on a table with [itex]μ_{s} = .15 [/itex] and [itex]μ_{k} = .1 [/itex]

    [itex]μ_{s}(F_{N}) = μ_{s}(mg)= .15(5)(10\frac{m}{s^{2}}) [/itex] =7.5N

    [itex]μ_{k}(F_{N}) = μ_{k}(mg)= .1(5)(10\frac{m}{s^{2}}) [/itex] = 5N


    The force applied to the block is represented by the function:

    [itex] f(x) = 7.5Sin(\frac{π}{10}x) [/itex]

    Where at time t = 5, static friction is broken and at time t = 7.67, equilibrium is reached, and at time t = 10, force is no longer applied.

    1_zpsgzhdx3cg.png


    Taking into account kinetic friction is 5N, I am imagining that from time t = 0 to t = 5, the 5kg block does not move even though there is an increasing force applied to it.

    Once the force applied to the block reaches 7.5N (which occurs at time t = 5), the Net Force on the block is [itex]Fnet = 7.5N - μ_{s}(F_{N}) = 7.5N - 5N = 2.5N[/itex]

    It just so happens that as static friction is broken at applied force of 7.5N, the force applied to the block begins to decrease.

    I imagine that a graph of the Net Force on the block would like exactly like this:

    6_zpse3tls3vd.png

    Which I would describe as piecewise function like this:

    [itex] y = 0 [/itex] from time t(0,5)
    [itex] 7.5Sin((\frac{π}{10})x) -5 [/itex] from time t(5,10)

    Now my goal right now is to be able to calculate the correct Impulse experienced by the block from time t(0,10).


    Question 1: Is there an impulse on the block between time 0s and 5s, which the block is stationary due a max 7.5N of static friction?

    Question 2: The impulse on the block between the time 5s and 7.672s seems to be 4N as described here:

    7_zpsfdcxqdvo.png

    However, to calculate the impulse between time 7.67s and 10s, I don't know how I should be considering the area under the curve. I guess it would be one of either two ways as illustrated here:

    8_zpsobfmynom.png

    Question 3:

    Does the block actually stop moving before 10 seconds?

    How can I know when this happens?

    Question 4:

    Can I only calculate an accurate impulse value once I first calculate the time at which the block is not longer moving? This may go back to question one: Does impulse on the block occur when a force is applied to the block even when the block remains stationary due to friction?

    Thank you
     
    Last edited: Oct 7, 2015
  2. jcsd
  3. Oct 7, 2015 #2
    It seems clear that what they want you to calculate is the impulse of the net force on the block. This is the integral of the net force dt.

    From t = 0 to t = 5 seconds, since the block is not accelerating, how is the static friction force related to the imposed force? (Note that the static friction force does not exceed that critical value of 7.5 N during the first 5 seconds)? What is the Net Force during the first 5 seconds?

    From t = 5 seconds to 10 seconds, ##Net Force = 7.5 \sin(\pi t/10) - 5##

    From this equation, what is the value of the externally imposed force and the Net Force at t = 7.67 sec. What does this tell you about the continuity of the net force at t = 7.67 sec?

    Given the answer to this question, what is the impulse of the net force from 5 sec to 10 sec? What is the impulse of the net force from 0 to 5 sec? What is the impulse of the net force from 0 to 10 sec?

    Chet
     
  4. Oct 7, 2015 #3
    Static force of friction pushes against the applied force with equal magnitude, but opposite direction from [itex]f_{a} = 0N [/itex] to [itex]f_{a} = 7.5N [/itex], producing a Net Force of 0N from t = 0 sec to t = 5 sec, which is why the block remains stationary until 7.5N. I was also considering the function from t = 0 sec to t = 5 sec to be [itex]f(t) = 7.5Sin(\frac{π}{10}t - 10)[/itex] so that net force at t = 5 sec would be 0N, but that describes static friction pushing back with a full 7.5N for the first 5 seconds, which would produce a net force in the negative direction, which is supposed to cause acceleration in the negative direction, which would make no sense. So y = 0 seemed to represent the net force of 0N.



    At t = 7.67 sec, Applied Force is 5N, Kinetic Friction is 5N, and Net Force is 0N. This is the only point in time when the block is in motion and in equilibrium, thus would begin traveling at a constant velocity if applied force remained at 5N. I believe the continuity at t = 7.67 sec remains continuous because there is no break in the function. At t = 5 sec, the graph jumps from Fnet = 0N to Fnet = 2.5N in what probably is an instant. Not sure if this is really possible or if there needs to be at least some time frame for the Net Force to build up from 0N to 2.5N, even if that time scale is something like .0001 sec.


    I am not sure what the impulse from t = 0 sec to t = 5 seconds should be. Does a stationary object have/produce an impulse or does an object have to have a change in velocity in order to have an impulse? The formula for impulse is Ft = mvf - mvi, so if the velocities are the same, then the impulse is 0. So if an object is stationary, then its impulse should be 0. So the impulse from t = 0 to t = 5 seconds should be 0, yes?

    The impulse from t = 5 sec to t = 7.67 sec is 4. But I calculated the impulse from t = 7.67 to t = 10 sec to be -5.18, worked out here:

    9_zps9u4hcyxd.png

    I am not 100 percent sure if something can have negative impulse, but my guess would be yes and that it would mean the object is slowing down. Perhaps that means for an object to begin moving and then become stationary, the total impulse would have to balance to zero. But here, total impulse is -1.18, which means there is more slowing down than speeding up, which can only mean a few things: If something is slowing down longer than it speeding up, then maybe the object is now moving in the negative direction. Or, the block already came to a complete stop somewhere between time interval t = 7.67 and t = 10 sec.
     
  5. Oct 7, 2015 #4
    Your first assessment was correct for the first 5 seconds:

    Applied external force = 7.5 sin (πt/10)

    Static frictional force = - 7.5 sin (πt/10)

    Net force = 0

    The static friction force is not equal to 7.5 N throughout the first 5 seconds. It is only equal to 7.5 N at t = 5 seconds. This happens to also be the critical force required for the block to start slipping, so static friction just releases at 5 sec.
    Our description of friction is only an idealization. So you are correct in this assessment.
    Yes. We already said that we are trying to determine the impulse of the Net Force, and the Net Force is zero during the first 5 sec.

    You've done very well in this assessment. A minor correction is that you used 7.6 sec in the first part of your calculation instead of 7.67. However, you are correct that, if integrated all the way out to 10 seconds, the impulse of the net force would be negative (-1.13 N-sec). This tells you that, at a certain time (close to 10 sec), when the impulse is zero, the block would come to a stop. The external force on the block would then be pointing in the negative direction. The question is whether the magnitude of this force would then be greater or equal to 7.5 N, so that static friction could again be overcome (of course, the static friction force after this time would be in the positive direction). The answer is, of course, that it wouldn't (since the maximum that the external force can be, according to our equation, is 7.5N). So, at the time that the block stops, the impulse of the net force would be zero, and it would stay zero out to 10 sec.

    Chet
     
  6. Oct 13, 2015 #5
    Thank you Chestermiller,

    To find the point on the t axis where the net impulse (net area under the curve) between 5s and 10s:

    [itex]\int^{t}_{5} 7.5sin(\frac{pi}{10}t) -5 [/itex] = 0

    [itex] (-\frac{75}{pi})cos(\frac{pi}{10}t) - 5t + 25 = 0 [/itex]

    To solve this for t, I used Newton's Method of Approximation:

    [itex]x_{c} ≈ x_{1} - \frac{f(x_{1})}{f'(x_{1})}[/itex]

    [itex] x_{c} ≈ \frac{ -\frac{75}{π}cos(\frac{π}{10}t) - 5t + 25 } {7.5sin(\frac{π}{10}t)} - 5 [/itex]

    First, I chose any number between 5 and 10, which was [itex] x_{1} = 6 [/itex], but that lead me to [itex] x_{c} = 5[/itex], which is another point in which the area under the curve = 0.

    I then realized I needed to choose any point between 7.67s and 10s,
    So I chose a [itex] x_{1} = 8 [/itex], which lead me to [itex] x_{c} = 9.76s[/itex]

    So going back to the formula for impulse, it can be shown that 0 impulse = 0 velocity

    [itex] F * t = mv_{f} - mv_{i} [/itex]

    [itex] 0 = mv_{f} - m(0) [/itex]

    [itex] 0 = mv_{f} [/itex]

    [itex] \frac{0}{m} = v_{f} [/itex]

    [itex] 0 = v_{f}[/itex]

    So the object stops moving at approximately t = 9.76s, which is .24 seconds before there is no more force applied to the object.
     
    Last edited: Oct 13, 2015
  7. Oct 13, 2015 #6
    Your application of Newton's approximation does not look correct to me. Please check it out. Also, see what you get if you choose your initial guess at t = 10s. This should be very accurate since it is very close to the actual solution, and the behavior of the function is nearly linear in this region.

    Chet
     
  8. Oct 14, 2015 #7

    Thank you Chet,

    The mistake is corrected.

    To find the value on the t axis where the net impulse (net area under the curve) between 5s and 10s:

    [itex]\int^{t}_{5} 7.5sin(\frac{pi}{10}t) -5 [/itex] = 0

    [itex] (-\frac{75}{pi})cos(\frac{pi}{10}t) - 5t + 25 = 0 [/itex]

    To solve this for t, I used Newton's Method of Approximation:

    [itex]t_{c} ≈ t_{1} - \frac{f(t_{1})}{f'(t_{1})}[/itex]

    [itex] t_{c} ≈ t_{1} - \frac{ -\frac{75}{π}cos(\frac{π}{10}t) - 5t + 25 } {7.5sin(\frac{π}{10}t) - 5} [/itex]

    Here is the work, which I meant to include in my previous post:

    390c0d13-b25a-4763-b606-4288357b137f_zps71rcvejt.jpg




    In approximating the t value where impulse equals zero by choosing [itex] t_{1} = 10 [/itex]

    I get: [itex] t_{c} = 9.76122 [/itex] and this is calculated in fewer steps than when I chose to start with [itex] t_{1} = 8 [/itex]!



    Also, if I were to incorporate the advice from another thread:

    https://www.physicsforums.com/threads/friction-force-has-different-effect-than-applied-force.835346/#post-5246464


    In post #10, CWatters says:


    CWatters explained:




    If I start out with [itex]f(x) = 7.5sin(\frac{pi*t}{10} – 5 [/itex]

    Then since [f = ma] = [itex] \frac{f}{m} = a [/itex] = [itex] \frac{d}{dt}\frac{f}{m} = a(t)[/itex]

    [itex]a = \frac{7.5sin(\frac{pi*t}{10}) – 5}{5} [/itex]

    [itex]a = \frac{7.5Sin(\frac{pi*t}{10})} {5} - \frac{5}{5} [/itex]


    [itex]a = \frac{f}{m} = \frac{7.5}{5}sin(\frac{pi*t}{10}) - 1 [/itex]


    [itex]\int^{t}_{5} a = v = \int \frac{7.5}{5} Sin(\frac{pi*t}{10}) - \int 1 [/itex]

    [itex] v(t) = -\frac{75}{5pi}cos(\frac{pi*t}{10})]^{t}_{5} [/itex]

    [itex]v(t) = -\frac{75}{5pi}cos(\frac{pi*t}{10}) - t + 5[/itex]

    and at t = 9.76122s

    v(9.76122) = 0.000000593s

    This seems to work as well.

    So velocity of an object can be determined 2 ways, which actually seems to be the same way in a different order.

    Method 1:
    If I start with a function for Net Force: [itex]f(x) = 7.5sin(\frac{pi*t}{10}) – 5[/itex]

    I can find the impulse from time = 5s to some time = t, say t = 7.67,

    [itex] \int^{7.67}_{5} F(t) dt = I(t) = \int^{7.67}_{5} 7.5sin(\frac{pi}{10}t) -5 [/itex] = 4.4

    [itex] [I = F*t = mv_{f} - mv_{i} ][/itex] = [itex] [4.4 = (5kg)v_{f} - (5kg)(0m/s)] [/itex]

    [itex] [4.4 = (5kg)v_{f}] [/itex] = [itex] [ \frac{4.4}{5} = v_{f} [/itex] = [itex].88 \frac{m}{s^{2}}] [/itex]

    so at v(7.67s) = [itex].88\frac{m}{s^{2}}[/itex]

    Method 2:

    Again, if I start out with a function for Net Force: [itex]f(x) = 7.5sin(\frac{pi*t}{10}) – 5 [/itex]

    since F = ma = [itex][ \frac{F}{m} ] [/itex] = a

    [itex] \frac{7.5sin(\frac{pi*t}{10}) – 5}{5} [/itex] = a

    [itex]= \frac{f}{m} = \frac{7.5}{5}sin(\frac{pi*t}{10}) - 1 [/itex] = a

    and [itex] \int^{t}_{5} a(t)dt [/itex] = v(t) , and

    [itex] v(t) -\frac{75}{5pi}cos(\frac{pi*t}{10}) - t + 5 [/itex]

    [itex] v(7.67) = -\frac{75}{5pi}cos(\frac{pi*7.67}{10}) - 7.67 + 5 = 32.32[/itex]


    Question:

    How come for method 1, the velocity arrived at was v(7.67) = .88 m/s
    and for method 2, the velocity arrived at was v(7.67) = 32.32 m/s ?


    Should this be what's occurring here:
    [itex][ \int \frac{F(t)}{m} dt = v(t) ][/itex] = [itex][ \frac{I(t)}{m} = v(t)] [/itex] ?

    Then how come v(t)≠v(t) ?
     
    Last edited: Oct 14, 2015
  9. Oct 14, 2015 #8
    In your equation for f', you are missing a -5.

    As far as your evaluations of the velocity at 7.67s are concerned, you need to do the arithmetic correctly.

    Chet
     
  10. Oct 15, 2015 #9
    Thank you Chet,

    I believe I fixed the -5 that you are referring to and then accounted for it in the Newton's Method approximation calculation. The correction is highlighted in blue in my previous post, but I had not made the correction on paper at the time of posting the image of my work on paper. I have since made the correction on paper as well.

    With regard to arriving at two different velocity values, I reviewed my arithmetic a few times and finally found that I was not entering the problem correctly into my calculator. Apparently, 75/5pi returns a different result than 75/(5pi). I arrived at the correct answer of approximately .8... for both methods.

    Thank you

    By the way, I have another question that extends this question to that of energy.

    I would like to know if there is a way to convert a Force vs. Time function into a Force vs. Distance function. For instance, if I know Force as a function of time and determine its impulse and can determine velocity at a given time, is it then possible to somehow create a new function such that the area under the curve will yield a value for energy?
     
    Last edited: Oct 15, 2015
  11. Oct 15, 2015 #10
    $$m\frac{dv}{dt}=F$$
    $$dt=\frac{dx}{v}$$
    So,
    $$mv\frac{dv}{dx}=F$$
    $$\int{Fdx}=\frac{1}{2}mv^2=KE$$
    Since you've already calculated the velocity, you know the amount of kinetic energy at time t. This is equal to the net work done on the block.

    Chet
     
  12. Oct 15, 2015 #11
    Thank you Chet,

    Before incorporating calculus, may I consider an object moving at a constant acceleration with a linear velocity?

    Suppose an object has velocity v(t) = t and is evaluated from 0 to 10 seconds.

    Then I believe that I may have correctly produced a Force vs. Time graph for the object:

    Force%20vs%20Time_zpse6i3ram6.jpg


    Then, I believe this may be a Force vs. Distance graph for the object:


    Force%20vs%20Distance_zpsgbnalwxv.jpg

    The issue I'm running into is, that since this graph is a Force vs. Distance graph, the function is in terms of distance, F(d) = ( something)

    The problem is that I don't know what the "something" is.

    Since the Force should be a constant 5N, I'm trying to come up with some form of (mass)(acceleration) that somehow reduces to a constant of 5.

    If force is constant 5N over time, then force should be constant 5N over a distance. So in both graphs, shouldn't the force be a constant:

    F(t) = 5, and F(d) = 5?

    How might I make F(d) = m(a) = 5?


    Thank you
     
  13. Oct 15, 2015 #12
    I'm really struggling to understand what you're asking (incidentally, the area under the F vs d graph is 250 Nm = 250 J).
     
  14. Oct 16, 2015 #13
    Oops, the area under the curve is definitely 250J! Thank you.


    With regard to the confusion I'm experiencing on representing the force function, let me back up a second.

    Suppose I have a 5kg object travelling at a constant velocity of 10m/s for 10 seconds.

    [itex]F(t) = ma(t) = m(\frac{Δv}{Δt}) = m(\frac{0}{Δt}) = 0N[/itex]

    However, if I write F(t) = ma like this:

    [itex]F(t) = ma(t) = m(\frac{Δv}{Δt}) = m(\frac{Δv}{(\frac{Δd}{Δv})}) = \frac{m(Δv)^{2}}{d} = \frac{m(Δv)^{2}}{( Δv )( Δt )} = \frac{5kg(10)^{2}}{(10)(10)} = 5N[/itex]

    I know the Net Force must be 0N because the velocity is constant, yet I'm not sure why rearranging the F = ma causes the Force to be 5N, as though there is a constant acceleration instead of a constant velocity. To be arriving at a constant, there must be something I'm not accounting for when rearranging f = ma.
     
  15. Oct 16, 2015 #14
    You already said that Δv = 0, so, in your second example F = 0.
     
  16. Oct 16, 2015 #15
    Thank you, of course Δv = 0m/s ≠ 10m/s. I should have seen that much sooner. I was actually in the process revising my post to make the correction, but you beat me to it.

    However, when going to the next level where an object is travelling at a constant acceleration of a = 1, I have not been able to account for the different values of 5N and 10N.

    Suppose an object is travelling at a velocity of v(t) = t

    [itex]F(t) = ma(t) = m(\frac{Δv}{Δt}) = m(\frac{10}{10}) = 5N[/itex]

    However,

    [itex]F(t) = ma(t) = m(\frac{Δv}{Δt}) = m(\frac{Δv}{(\frac{Δd}{Δv})}) = \frac{m(Δv)^{2}}{d} = \frac{m(Δv)^{2}}{\frac{1}{2}( Δv )( Δt )} = \frac{5kg(10)^{2}}{\frac{1}{2}(10)(10)} = 10N[/itex]

    Note, I'm supposing distance equals (1/2)(base)(height)

    What misconception might account for the value of 10N that I'm arriving at?
     
    Last edited: Oct 16, 2015
  17. Oct 16, 2015 #16
    [itex]F(t) = ma(t) = m(\frac{Δv}{Δt}) = m(\frac{Δv}{(\frac{Δd}{Δv})}) = \frac{m(Δv)^{2}}{d} = \frac{m(Δv)^{2}}{\frac{1}{2}( Δv )( Δt )} = m2 \frac{v}{t} = 2m\frac{\frac{d}{t}}{t}= 2m\frac{d}{t^{2}}= 2m\frac{\frac{1}{2}t^{2}}{t^{2}}= [/itex] m !!

    Got it! Don't know why I have to make 2 substitutions for distance. I figured one would be enough, but it works.
     
  18. Oct 17, 2015 #17
    [itex]F = ma = m\frac{dv}{dt} = m\frac{dv}{\frac{dx}{v}} = mv\frac{dv}{dx} [/itex]

    [itex]\int F = \int ma = \int m\left ( \frac{F}{m} \right )= \int m \frac{dv}{dt} = \int m\frac{d}{dt}\left ( \int \frac{F}{m} \right )= \int_{5}^{9.76}m\frac{d}{dt}\left ( \int \frac{7.5Sin\left ( \frac{\pi *t}{10}-5 \right )}{m} \right ) = \int_{5}^{9.76}5\frac{d}{dt}\left ( \int \frac{7.5Sin\left ( \frac{\pi *t}{10}-5 \right )}{5} \right ) [/itex] ?

    This doesn't seem right because I am getting every number under the sun beside 0. I'm assuming that since the velocity at 5 seconds is 0 and the velocity at 9.76 seconds is 0, then the change in kinetic energy should be 0.
     
    Last edited: Oct 17, 2015
  19. Oct 17, 2015 #18
    I have no idea what you are doing. You already calculated the velocity in previous posts and showed that it is zero at 9.76 seconds.

    Chet
     
  20. Oct 18, 2015 #19
    Hi Chet,

    Maybe I can better describe what I'm looking to find with the following image:

    impulse%20to%20energy_zpsxrey21m8.jpg


    I'm trying to start with a function of Force in terms of time, which we've been working with, f(t) = 7.5Sin(\frac{pi*t}{10}) - 5, and then from this function, come up with a function of Force in terms of friction. In other words, I am trying to write a function of force in terms of distance from an already known function of force in terms of time, if it is possible to convert.

    For instance, at a distance of say, 10 meters, what force is being applied? At 20 meters, what force is being applied?
     
  21. Oct 18, 2015 #20
    You already know F(t) and v(t). Step 1: Integrate v(t) at get d(t). So now you have equations for the force and distance, expressed parametrically in terms of time. Step 2: Just solve for t in one of the equations and substitute it into the other equation. Or make a table with 3 columns, with t in the first column, f(t) in the second column, and d(t) in the third column. Then just plot the second column as a function of the third column.

    Chet
     
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