Strings and Springs?

Question

Your friend's 12.6 g graduation tassel hangs on a string from his rear-view mirror. a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car. Explain. b) If the tassel hangs at an angle of 6.20 degrees relative to the vertical, what is the acceleration of the car?

Answer 1

a) With the car at rest, the only forces acting on the tassel are
1. its weight, downwards;
2. string tension, upwards.

Both forces mutually cancel each other, and, since no horizontal forces are involved, the tassel is therefore at rest.

But then the car accelerates; in order for the tassel to remain inside the car, it has to accelerate forward as well. The only physical means for this to happen, is by being pulled by the string.

A vertical string can only pull upwards. However, an slanting string can provide an horizontal pull at the same time it pulls upwards. The slant angle is a measure of the relative magnitudes of the forces involved.

b) By the same token, as these forces act upon the same mass, the slant angle is also a measure of the relative magnitude of the vertical and horizontal components of acceleration imparted to the tassel. To be more specific, the relative accelerations are proportional to the vertical and horizontal distances into which the slanting string can be resolved.

The tangent (trigonometric function) relates both distances with the angle they form: tan θ = x / y = a / g, where a is the (horizontal) car acceleration, and g the acceleration of gravity. (Remember, θ is the angle the string makes with the VERTICAL).

Thus, a = g tan θ = 0.1086 g (about 1.07 m/s²).

Answer 2

a. Since it is accelerating, there is a constant change of velocity. this forces the tassel back. Constant velocity change.

b. (sin6.2)/9.8 = acceleration of car in meters/sec squared

= 1.0583937 Meters/sec (squared)