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

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. Maybe simple demo, add twice as much mass, get twice the stretching before showing the slide. Also make note that this is NOT as big a deal as Newton’s laws. Just an example of a contact force.

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. Maybe simple demo, add twice as much mass, get twice the stretching before showing the slide. Also make note that this is NOT as big a deal as Newton’s laws. Just an example of a contact force. FBD: Fs Newton’s Second Law: F = ma Fs- Fg = 0 Fs = Fg k x = m g k = 612 N/m y x Fg

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

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: FT Newton’s Second Law: F = ma FT - Fg = 0 FT = Fg FT = m g FT= 440 N y x Fg

Summary Contact Force: Spring Contact Force: Tension 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

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 y x  Ff Fg FT

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. y x FN  FTcos Ff Fg FTsin

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: y x F = ma FTcos - Ff = 0 FN - FTsin - Fg = 0 FN FTcos Ff y-direction: Fg FTsin

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 FN - FTsin - Fg = 0 FN = FTsin + mg FTcos = (FTsin + mg) FT = mg / (cos - sin)

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

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 y x  Fg

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: y x F = ma FTcos - Ff = 0 FN - FTsin - Fg = 0 FN FTsin Ff y-direction: FTcos Fg

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: y x F = ma FTcos - Ff = 0 FN - FTsin - Fg = 0 FN FTcos Ff y-direction: Fg FTsin