For the first exercise, a uniform meter stick of mass 167 gram is suspended as s
ID: 1362421 • Letter: F
Question
For the first exercise, a uniform meter stick of mass 167 gram is suspended as shown below. Two masses are hung at the positions shown. If the mass of m^1 is 169 gram, what is the mass of m^2 so that the meter stick is in static equilibrium? 15.337 g Submit Answer Incorrect. Tries 1/10 Previous Tries Next mass nr^ = 111 gram is hung as shown below. You have mass, m_2 = 181 gram. Where should m_2 be hung so that the meter stick is in static equilibrium? Recall that the mass of the meter stick is 167 gram. Click on the meter stick the location where m_2 should be hung. Then click on the "Submit Answer" button.Explanation / Answer
given,
length of stick = 1 m
m1 = 169 gm
position of fulcrum = 36 cm
since the system id in equilibrium the moment if weight about fulcrum must be equal
m1 * d1 = m2 = d2
169 * (36 - 13) = m2 * (65 - 36)
m2 = 134.0345 gm