A pear and a papaya are placed 12 meters apart from each other. The pear is 0.005 kg and the papaya is 0.008
kg. Calculate the force of gravitation between them.

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).

The system is released from rest with the cable taut, and the homogeneous cylinder does not slip on the rough incline. determine the angular acceleration of the cylinder and the minimum coeffi cient s of friction for which the cylinder will not slip.

A20 cm long spring is attached to a wall. the spring stretches to a length of 22 cm when you pull on it with a force of 100 n. what is the spring constant