University:
Jackson College
Course:
PHY 231 | College Physics I
Academic year:

2023
Views:

397

Pages:

4
Author:

bodeur1vli

**Formulas:** ΣFx = 0 ΣFy = 0 ΣF = 0 **1. F1, F2, and F3 are the weights of three objects suspended by pulleys, as illustrated. Assuming the pulleys in this system are frictionless and weightless and that the magnitude of the third weight F3 is 12 N, what are the magnitudes of F1 and F2?** [Image of a pulley system with three weights] F1 = 12 * cos 40° F1 = 7.7 N F2 = 12 * cos 50° F2 = 9.2 N **2. An object is suspended on a frictionless inclined plane by a rope parallel to the incline, as illustrated above. If the angle of the incline is 25.0° and the tension on the rope is 5 000 N, what is the weight of the object?** cos 65° = 5000 / Fg Fg = 5000 / cos 65° Fg = 1.18 x 10^4 N **Castle Rock Research** **Page 38** **Lesson 1 - Practice Exercises** 1. A 20.0 N child sitting on a playground swing is being pushed by her father. When the swing rope makes an angle of 27.0° to the vertical, what is the force exerted by her father? What is the magnitude of tension in the rope, Ft? [Image of a playground swing with a child and angles labeled] tan 27° = F / 20 F = 10.2 N cos 27° = 20 / Ft Ft = 22.4 N 2. Two ropes are attached to a 75.0 N object as illustrated. Find the magnitude of the tension forces, F_T1 and F_T2, in the ropes. [Image of an object suspended by two ropes at angles] sin 68° = F_T1 / 75 F_T1 = 15.4 N sin 22° = F_T2 / 75 F_T2 = 82.0 N 3. A 15 kg object rests on a table. A cord is attached to this object and also to a wall. Another object is hung from this cord as shown above. If the coefficient of friction between the 15 kg object and the table is 0.27, what is the maximum mass that can be hung? **EQUILIBRIUM - Practice Exercises** 6. A 735 N mountain climber is rappelling down the face of a vertical cliff. If the rope makes an angle of 12° with the vertical face, what is the magnitude of tension in the rope? [Image of a mountain climber rappelling down a cliff] cos12° = 735 / Ft Ft = 735 / cos12° Ft = 7.5 x 10² N 7. In the static arrangement shown above, find F₁ and F₂. [Image of a static arrangement with two forces and an angle] tan30° = F₂ / 96 F₂ = 96 tan30° F₂ = 55 N sin60° = 96 / F₁ F₁ = 96 / sin60° F₁ = 1.1 x 10² N 8. In the static arrangement shown above, find F₁ and F₂. [Image of a static arrangement with two forces and angles] 170 - 60 - 30 = 80 Use side lengths to find angles. tan θ = 30 / 50 θ = 26.57° sin26.57° = 55 / F₁ F₁ = 55 / sin26.57° F₁ = 1.2 x 10² N