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1. FT: Force of Tension Fg : Force of Gravity or “Weight” FT Fg Force Vector Diagram #1: An elevator is ascending or descending and you want to know what is the tension in the rope.

2. Speeding Up Slowing Down Going Up Acceleration Acceleration Going Down Acceleration Acceleration First you will need to find the direction of the acceleration.To do so use the chart below:

3. FT a Fg • When the acceleration is upwards use the following equation… • Fy = FT – Fg = ma • åFy is the “sum of the forces” in the y direction • FT is the force of tension in Newtons • Fg is the weight or force of gravity in Newtons • m is the mass of the object in kilograms • a is the acceleration of the object in m/s2

4. FT a Fg • When the acceleration is downwards use the following equation… • Fy = Fg – FT = ma • åFy is the “sum of the forces” in the y direction • FT is the force of tension in Newtons • Fg is the weight or force of gravity in Newtons • m is the mass of the object in kilograms • a is the acceleration of the object in m/s2

5. Other things you need to remember: To find the mass of the object when given the weight in Newtons, divide by9.8 m/s2. To find the weight of the object when given the mass in kg’s, multiply by9.8 m/s2.

6. FT = ? a = 3m/s2 ** Fg= 200 x 9.8 = 1960N ** (use the acceleration table to decide on the direction of this arrow) Now let’s see an example… Example 1: There is an elevator that is going upwards and slowing down at a rate of 3 m/s2. It masses 200 kg’s. What is the tension in the cable from which it is hanging?

7. Plug what you know into the correct equation and solve for the object’s force of tension: Use the following equation: åFy = Fg – FT = ma Where: Fg = 200 kg x 9.8 m/s2 = 1960 N FT = ? m = 200 kg a = 3 m/s2 åFy= 1960 – FT = 200 x 3 – FT = 600 – 1960 FT = 1360 N

8. FT = ? a = 3m/s2 ** Fg= 200 x 9.8 = 1960N ** (use the acceleration table to decide on the direction of this arrow) Now let’s see another example… Example 2: There is an elevator that is going upwards and speeding up at a rate of 3 m/s2. It masses 200 kg’s. What is the tension in the cable from which it is hanging?

9. Plug what you know into the correct equation and solve for the object’s force of tension: Use the following equation: åFy = FT – Fg = ma Where: Fg = 200 kg x 9.8 m/s2 = 1960 N FT = ? m = 200 kg a = 3 m/s2 åFy= FT – 1960 = 200 x 3 FT = 600 + 1960 FT = 2560 N

10. Real life applications… • Forces in the y direction are used to move elevators up and down. • Forces in the y direction can be used to hoist things up and down from a tree house. • Forces applied in the y direction make a yo-yo go up and down. • Forces are applied in the y direction to pick something up, like the body builder below

11. The End Now try numbers 1 and 2 of your HW.Remember: you MUST draw the force-vector-diagram for EACH problem, and show the equation and ALL work for EACH problem, if you want to get full credit on your HW!!