A meter rule is balanced at the 70 cm mark by placing a 200-g mass at the 80 cm
ID: 1913747 • Letter: A
Question
A meter rule is balanced at the 70 cm mark by placing a 200-g mass at the 80 cm mark. At what point will the ruler balance if the 200-g mass is moved to the 90 cm mark? Two blocks of masses 15kg and 25kg are on a double inclined plane connected by a rope. What is the tension on the rope connecting the two masses? What is the acceleration of the two-mass system? A crate weighing 9800 N is pulled up a 37o incline by a pulley system. If the coefficient of kinetic friction between the crate and the plane is 0.75 N, what is the total force that opposes its motion up the plane?Explanation / Answer
At the 58.18 cm (approx at the 52 cm) mark. The moment on the forces on both sides of the pivot should be equal. The forces are weight of the 130 gram and 20 gm mass on the left side (if the ruler started from 0 on the left to 100 at the right tip) and 500 gram on the right. The distances FROM THE PIVOT are 25 cm for the 130 gram, 42 cm (50-8) for the 20 gm and X for the 500 gm. So (130G)25 + (20G)42 = (500G)X, giving a value of X as 8.18, but this is from the pivot to the right so it should be placed at the 50+8.18 = 58.18 mark. P.S.: the "G" stands for acceleration due to gravity Use consistent units ALWAYS, here we use grams for mass and cm for length, a better idea is to convert these to SI units, more complex problems might require such standard units be used.