Forces - 2 - Weight & More Equilibrium

University:
High School
Course:
High School Physics
Academic year:

2024
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14
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2.3 Weight: Weight describes the force of gravity on an object. ◼ It is a force, measured in Newtons ◼ g = 9.80 m/s2 (or N/kg) [downwards] (on earth) ◼ But if the object is on an incline: ◼ Why mg sin θ ? Ex. A 100.0kg object slides down a plane inclined at 40.0o. What is the magnitude of the force that causes the motion? 40o F=? Ex. A TV screen is suspended from a ceiling. Given both the angles and tensions in the wires, find the weight & mass of the screen. T=115N 65.0o T=115N 65.0o Weight = Net Vertical Force 115N Fg = mg 104.225N 65o 208.450N = m (9.80N/kg) So, mass = 21.2704 kg = 21.3 kg T=115N 65.0o Vertical component of the tension in wires is equal to weight of TV same for both wires… So, weight = 208 N T=115N 65.0o Ex. Determine the tension in each wire suspending speakers, with a mass of 40.0kg, over a stage. 35.0o The angles are the same, so each rope supports half the load of 40.0kg ? 35.0o 20.0kg 196 N F = mg F = (20.0)(9.80) F = 196 N vertical force 6 Ex. Two wires are used to suspend a 50.0kg sign, as shown. T1 = 386.0N T2 = ? 65.0o 40.7o Coffee What is the tension on the other wire? (assume the system is in equilibium) Problem: Different Angles!!! 7 ? sin 65 = 386 349.835N vertically F = mg o Other wire = difference 140.165N vertically F = (50kg)(9.80N/kg) So weight sign = 490N Check…horizontal components should be equal but in opposite direction!!! 140.165 sin 40.7 = ? o 163.131N horizontally for both Tension on wire = 214.945N = 215 N 8 Summary: • • • • System in equilibrium -> Net force = 0 N (x & y) Horizontal forces balanced -> x-comps of tension equal and opposite (left = right) Vertical forces balanced -> y-comps of tension (up) must equal force on the sign (down). Each y-comp will be different due to the different angles, but sum of y’s equals weight! 9 Ex. Two wires are used to suspend a 50.0kg sign, as shown. What is the unknown tension on wire 1? T2 = 200.0N T1 = ? 55.0o PW 200.0N F = 490.0N 490.000 – 163.830 = 326.170N 345.7N 163.830N 55.0o 114.715N FT = 346 N 114.715N 10 Ex. Two wires are used to suspend a 100.0kg sign, as shown. T1 = ? 75.0o T2 = ? 25.0o STARBUCKS What is the tension on each wire? (assume that the system is in equilibrium) 11 Ex. Two wires are used to suspend a 80.0kg sign, as shown. T1 = ? 68.0o T2 = ? 18.0o STARBUCKS What is the tension on each wire? (assume that the system is in equilibrium) 12 Ex. Use SINE LAW!!! T1 = ? T2 = ? T1 980 oT2 = 1565o = sin 65.0 75sin 15.025o sin 100.0 o T2 25o STARBUCKS 25o 75o T1 = 901.883N T2 = 257.556N Fg = 980N T1 = 902 N T1 o 75 T2 = 258 N 13 Quantum HW: pg 298 #2, #3 pg 302 #6 14