Physics Friday 149

For a block sliding down an inclined surface with angle of incline θ and with coefficient of kinetic friction μ. If we consider the forces, we see that the block can slide down the incline with constant velocity when the net forces are zero. Balancing forces perpendicular to the plane, the normal force is thus N=mg\cos\theta. The kinetic friction is thus f=\mu{N}=\mu{mg}\cos\theta, while the component of gravity parallel to the plane is thus mg\sin\theta. We see that these cancel when
\mu\cos\theta=\sin\theta, or when \mu=\tan\theta.

Now, suppose we have a large incline with \mu=\tan\theta, and we start our block sliding with a velocity v0 in a horizontal direction; that is to say, along the plane in a direction perpendicular to the direction of the slope. What, then, will be the speed a long time later?


Let v be the total speed of the block, and let vy be the velocity component in the downslope direction. Now, we have two forces parallel to the plane; the friction force, with magnitude
f=\mu{N}=\mu{mg}\cos\theta=mg\sin\theta
directed opposite to the motion of the block, and the component of gravity, also of magnitude mg\sin\theta, directed downslope.
The former produces an acceleration of magnitude g\sin\theta opposite the direction of motion, while the gravity gives the same acceleration g\sin\theta in the downslope direction; thus
\frac{dv}{dt}=-g\sin\theta, and \frac{dv_y}{dt}=g\sin\theta, so that
\frac{dv}{dt}+\frac{dv_y}{dt}=0, and thus
v+v_y=C, for C some constant.
Now, at time t=0, our motion is purely horizontal, and so vy=0, v=v0, and so
C=v0.
Now, after a long time, the horizontal component of the velocity will be effectively zero, so that the velocity is entirely downslope. Then v=vy, and so v+v_y=v_0 becomes
2v=v_0,
and so the speed after a long time is
v=v_y=\frac12v_0, half the initial speed; now in the downslope direction.

Advertisements

Tags: , , ,

2 Responses to “Physics Friday 149”

  1. Dr.D Says:

    In the definition of vo, what on earth do you mean by the phrase, “along the plane in a direction perpendicular to the direction of the slope?” That is about as muddy as any definition I have ever seen. I have no idea what you are saying vo is, so the rest of the problem is meaningless to me.

  2. twistedone151 Says:

    When we usually do this problem, we usually only consider motion directly up or down the plane, such as with the ball rolling down the plane here. I mean motion in the direction on the surface of the plane which is horizontal, neither uphill nor downhill. In the case of a diagram like in the above problem, it would be motion directly into, or directly out of, the plane of the diagram. If you draw the plane from above, and indicate the direction of the slope of the plane (the direction straight uphill or downhill, as it were), then this direction is along the line in the plane perpendicular to this direction (line of constant elevation).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: