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Snowboarding

Earlier, we analyzed the situation of a downhill skier moving at constant velocity to determine the coefficient of kinetic friction. Now let’s do a similar analysis to determine acceleration. The snowboarder of [link] glides down a slope that is inclined at θ = 13 0 to the horizontal. The coefficient of kinetic friction between the board and the snow is μ k = 0.20 . What is the acceleration of the snowboarder?

Figure (a) shows an illustration of a snowboarder on a slope inclined at 13 degrees above the horizontal. An arrow indicates an acceleration, a, downslope. Figure (b) shows the free body diagram of the snowboarder. The forces are  m g cosine 13 degrees into the slope, perpendicular to the surface, N, out of the slope, perpendicular to the surface, m g sine 13 degrees downslope parallel to the surface and mu sub k times N, upslope parallel to the surface.
(a) A snowboarder glides down a slope inclined at 13° to the horizontal. (b) The free-body diagram of the snowboarder.

Strategy

The forces acting on the snowboarder are her weight and the contact force of the slope, which has a component normal to the incline and a component along the incline (force of kinetic friction). Because she moves along the slope, the most convenient reference frame for analyzing her motion is one with the x -axis along and the y -axis perpendicular to the incline. In this frame, both the normal and the frictional forces lie along coordinate axes, the components of the weight are m g sin θ along the slope and m g cos θ at right angles into the slope , and the only acceleration is along the x -axis ( a y = 0 ) .

Solution

We can now apply Newton’s second law to the snowboarder:

F x = m a x F y = m a y m g sin θ μ k N = m a x N m g cos θ = m ( 0 ) .

From the second equation, N = m g cos θ . Upon substituting this into the first equation, we find

a x = g ( sin θ μ k cos θ ) = g ( sin 13 ° 0.20 cos 13 ° ) = 0.29 m/s 2 .

Significance

Notice from this equation that if θ is small enough or μ k is large enough, a x is negative, that is, the snowboarder slows down.

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Check Your Understanding The snowboarder is now moving down a hill with incline 10.0 ° . What is the skier’s acceleration?

−0.23 m/s 2 ; the negative sign indicates that the snowboarder is slowing down.

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Summary

  • Friction is a contact force that opposes the motion or attempted motion between two systems. Simple friction is proportional to the normal force N supporting the two systems.
  • The magnitude of static friction force between two materials stationary relative to each other is determined using the coefficient of static friction, which depends on both materials.
  • The kinetic friction force between two materials moving relative to each other is determined using the coefficient of kinetic friction, which also depends on both materials and is always less than the coefficient of static friction.

Conceptual questions

The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.

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When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction.

If you do not let up on the brake pedal, the car’s wheels will lock so that they are not rolling; sliding friction is now involved and the sudden change (due to the larger force of static friction) causes the jerk.

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When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular, explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)

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Practice Key Terms 3

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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