University of Wisconsin Green Bay

A 15 kg block rests on an inclined plane. The plane makes an angle of 25o with the horizontal, and the coefficient of friction between the block and the plane is 0.13. The 15 kg block is tied to a second block (mass=38 kg) which hangs over the end of the inclined plane after the rope passes over an ideal pulley. What is the acceleration of each of the two blocks, and what is the tension in the rope?

  • In this problem, you are asked to relate motion (the acceleration of the two blocks) to force (tension in the rope, friction). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem.

    In addition, note that you must treat the blocks as separate systems. You are asked to find the tension in the rope between them, and cannot answer that question without examining the interaction between them—in other words, the effect that each has on the other. Therefore, you will need to draw a picture and set up equations for each block individually.

  • Step 1

    Blocks Tied Together

    Your FBD for Block 1 is not yet finished, because mg has both x- and y- components. Continue down to step 2 when you are ready to continue.


    Step 2

    Blocks Tied Together

    In the final FBD drawn here, all forces on Block 1 are divided into components. The contribution each force makes in the x-direction (along the incline) is shown explicitly, as is the contribution each force makes in the y-direction. The FBD is now a visual representation of ∑F=ma in each direction.

    Tension Tension T +x T y X Y Fr Fr m1g m1g cos25^0 m1g sin25^0
  • Stacked Boxes with Friction

    The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed.

    For multiple object problems, you will always need the additional information summarized as Newton’s Third Law (the interaction between two objects is felt by both objects equally and in the opposite direction.) In this example, that understanding has already been used—the interaction between the two blocks comes through tension in the rope, and tension was given the same symbol for each. If you did not recognize that tension is the same throughout the rope as you drew the FBDs, that is ok. When you start to solve the equations, you will find that you have too many unknowns and you can use this understanding to reduce them at that point.

  • Step 1

    Figure 2 Equations

    One of the keys to successfully solving a multi-object problem algebraically is to keep track of the variables. I used different symbols for the masses of the two blocks because they are not the same, but I used the same symbol for acceleration because they move together. I also used the same symbol for tension on each block.

    At this point, you have two unsolved equations and two unknowns (a and T.). Scroll down to continue this solution.


    Step 2

    At this stage in the problem, we have two unknowns, a and T, and two unsolved equations:

    T – 370 N = -(38 kg)a

    T – 79 N = (15 kg) a

    One approach that always works is to solve one equation for one of the variables and substitute it into the other.

    T = 370 N – (38 kg)a from the first equation
    370 N – (38 kg)a - 79 N = (15 kg) a substituting into the second
    290 N = (38 kg + 15 kg)a
    5.5 m/s2=a

    Now that you have solved for one of the unknown variables, substitute it into either of the original equations to solve for the other variable. I will substitute it into the second equation.

    T – 79 N = (15 kg)(5.5 m/s2)
    T = 79 N + 83 N = 160 N

    Tension in the rope and acceleration of the blocks are the only information requested in this problem. No further mathematical solution is necessary.

    -a a gravity gravity gravity Fr Fr 17N 62N 130N 130N 130N 130N 130N 130N 130N

  • In this problem, we were asked to find the acceleration of two blocks tied together with a rope, and also to find the tension in the rope between them. The problem did not state for sure which direction the blocks are moving, or even if they are moving. Based on the relative masses of the blocks, we made the assumption that the hanging block accelerates downward and the sliding block upwards along the incline.

    With that assumption, our solution is:

    1.) The hanging block accelerates downward with a=5.5 m/s2 and the block on the incline accelerates upward along the incline, also with a=5.5 m/s2. This is just a little bit more than half the acceleration that the hanging block would have due to gravity alone, which makes sense. You would expect it to have a lower acceleration because of the upward tension in the rope due to the pull of the other block.

    2.) The tension in the rope is 160 N. This value is about the same as the gravitational force on a 16 kg mass, again a number that makes sense. The 38 kg block is accelerating downward, and so the tension in the rope does not fully support it against gravity. Therefore, you know the tension in the rope must be less than (38 kg)g or less than 370 N.

    Therefore, we correctly chose the direction of acceleration and friction.