Question

A: An inverted "V" is made of uniform boards and weighs 191 N. Each side has the same length and makes a 30.0° angle with the vertical, as the drawing shows. Find the magnitude of the static frictional force that acts on the lower end of each leg of the "V."

B: A small 0.542-kg object moves on a frictionless horizontal table in a circular path of radius 1.31 m. The angular speed is 6.33 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Answer 1

A consider the left board, which has a length & and a weight of (1910) = 95.5N - Let tv be the upward normal force exerted by the ground on the board. This torce balonces the weight, so fv = 95.50.. under equilibrium condition, &c=0 (95-50) Sin 30 (\) + fs (LCos 30) - Fv (I sin30) = 0 23.875N & fs Cos30 - Ty sin38 = 0 fs= (45.5%) sin30 - 23.875N_276568N) Co $38 from of Angular momentum. Conservation Iw= o'm lowo ~= (19) = w.- @ de centripetal acceleration (as) = Now, V=WR, SO, RAST - from e , solwing for R R = (m Rolls) = 0.542Kg (1.31m) 46-33 yadr)? 105N 0.609119622 m3 R= [0-847m