Lecture 02: Forces & 2-D Statics

1. Lecture 02: Forces & 2-D Statics • Springs • Tension • 2-D Statics Examples

2. Contact Force: Springs • Force exerted by a spring: Fspring= k*x • The greater the compression or extension of the spring, the greater the force. • The greater the “spring constant” (a characteristic of the spring), the greater the force.

3. Fs Fg y x Contact Force: Springs • Force exerted by a spring: Fspring= k*x • The greater the compression or extension of the spring, the greater the force. • The greater the “spring constant” (a characteristic of the spring), the greater the force. • Example:When a 5 kg mass is suspended from a spring, the spring stretches 8 cm. Determine the spring constant. FBD: Newton’s Second Law: F = ma Fs- Fg = 0 Fs = Fg k x = m g k = 612 N/m

4. Contact Force: Tension • Tension in an Ideal String: • Magnitude of tension is equal everywhere. • Direction is parallel to string (only pulls).

5. FT Fg y x Contact Force: Tension • Tension in an Ideal String: • Magnitude of tension is equal everywhere. • Direction is parallel to string (only pulls). • Example: Determine force applied to string to hold a 45 kg mass hanging over pulley FBD: Newton’s Second Law: F = ma FT - Fg = 0 FT = Fg FT = m g FT= 440 N

6. Summary • Contact Force: Spring • Can push or pull, force proportional to displacement • F = k x • Contact Force: Tension • Always Pulls, tension equal everywhere • Force parallel to string • Next, A Two Dimensional Example: • Choose coordinate system • Analysis of each direction is independent

7. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • We will choose a standard x,y coordinate system. • There are 4 forces in the FBD (Gravity, Normal, Friction, Tension). FN  Ff Fg FT

8. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • Tension is in both the x and y directions, so we must break it down into its x and y components. FN  FTcos Ff Fg FTsin

9. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • Now we will write Newton’s Second Law for each direction. • In each direction the acceleration is zero. x-direction: F = ma FTcos - Ff = 0 F = ma FN - FTsin - Fg = 0 FN FTcos y-direction: Ff Fg FTsin

10. Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • In the first equation, replace Ff with *FN. • Solve the second equation for FN, and replace Fg with m*g. • Now substitute the expression for FN into the first equation and solve. x-direction y-direction combine FTcos - Ff = 0 FTcos = FN FTcos = (FTsin + mg) FT = mg / (cos - sin) FN - FTsin - Fg = 0 FN = FTsin + mg

11. Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • Finally, substitute in the given numbers! (note: it is usually easier to wait until the very end to do this…) FT = mg / (cos - sin) FT = 798 N

12. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • We will choose a standard x,y coordinate system. • There are 4 forces in the FBD (Gravity, Normal, Friction, Tension). FN FT Ff  Fg

13. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • Now we will write Newton’s Second Law for each direction. • In each direction the acceleration is zero. x-direction: F = ma FTcos - Ff = 0 F = ma FN - FTsin - Fg = 0 FN FTsin y-direction: Ff FTcos Fg

14. y x Example: Force at Angle • A person is pulling a 150 kg refrigerator across a floor with mk= 0.4 at a constant speed. If she is pulling down at an angle of  = 25º with respect to the horizontal, what is the magnitude of the force she exerts on the refrigerator (the tension)? • Now we will write Newton’s Second Law for each direction. • In each direction the acceleration is zero. x-direction: F = ma FTcos - Ff = 0 F = ma FN - FTsin - Fg = 0 FN FTcos y-direction: Ff Fg FTsin