A simple pendulum has length L and period T. As it passes through its equilibrium position, the string is suddenly clamped at its mid-point. The period then becomes:

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.15 times a second. a tack is stuck in the tire at a distance of 0.373 m from the rotation axis. noting that for every rotation the tack travels one circumference (a) find the tack's tangential speed. (b) what is the tacks radial acceleration?

The astronauts on the space shuttle flights experienced an acceleration of 29 m/s2 (about 3 "g's") during lift-off. what upward force must the astronaut's seat apply to the astronaut in order to cause this acceleration? assume the astronaut's mass is 70 kg, and compute this force when the acceleration is near the earth's surface so their weight equals latex: mg m g .