University of Wisconsin Green Bay

A father is pulling his daughter uphill on a sled in slushy snow. The daughter has a mass of 22.7 kg; the sled has a mass of 3.18 kg. The coefficient of kinetic friction between the hill and the sled is 0.212 and the coefficient of static friction between the hill and the sled is 0.317. How hard does the father have to pull the sled in order to keep it moving at a constant velocity? The child holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5o with the horizontal.

  • In this problem, you are asked to relate motion (the sled moves up the hill at a constant velocity) to force (how hard the father pulls the sled). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem.

  • Step 1

    Object on an Incline

    Your FBD 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

    Object on an Incline

    In the final FBD drawn here, all forces 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.

    mg mg mg mg mg sin(15.5^0) mg cos(15.5^0) mg cos(15.5^0) Constant Velocity Spring Force Spring Force Gravity Gravity Friction Friction Tension Tension Tension
  • Object on an Incline

    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.

  • mg P n n Fr p

    The problem asks how hard the father needs to pull the sled, so no further steps are needed to solve the problem. Note that you do not need to think about whether friction is static or kinetic until you fill in for the frictional force, and the problem would be worked in exactly the same way if you were given force and asked for acceleration, or given force and acceleration and asked for the coefficient of friction.

    It does not matter whether you begin with the x- or the y-equation. If you solve for x first, you will need to pause in your solution to obtain a value for n from the y-equation.

  • In this problem, the child and sled are on a slight hill. Therefore, gravity does not act directly into the surface. This is seen in the figure with two components of gravity—one into the hill and one along it. It is seen in the solution because the normal force of 245 N (the supporting force of the hill) is less than the weight of the child and the sled (25.9 kg x 9.81 m/s2=254 N). You expect the two numbers to be close, which they are, because the hill is only at a slight incline.

    In the y-direction, the velocity of the box is also constant (the box remains at rest.) Therefore, the upward normal force from the floor exactly balances the downward pull of gravity.

    Both the equation and the figure represent the fact that the father has to balance two forces in order to keep his daughter moving at a constant velocity. He needs to balance a little bit against gravity (but not too much because the incline isn’t very steep) and he also needs to balance the force of friction that acts to slow the sled. The final answer of 120 N is less than the weight of the child and sled (254 N as seen above), and so the answer should also agree with your experience that less force is required to pull something up a snowy incline than to lift it.