Unit 5 Notes

∑F = ma

∑Fy = Upward Forces – Downward Forces = may

∑Fx = Forces to Right – Forces to Left = max

(∑ means the summation)

∑F is the net force.

∑F = Fnet

∑F = ma

*Keep in mind that a positive net force means acceleration in the positive direction

*A negative net force means acceleration in the negative direction

Example:A 50kg person in an elevator accelerates upward at 2 m/s2. Solve for the normal force.

∑Fy = FN + Fg =maFN + (50kg * -9.8N/kg) = 50kg * 2 m/s2

FN – 490N = 100NFN = 590N

Example 2A 70 kg para-sailer gets pulled by a cable that makes a 45 degree angle with horizontal. If the person’s acceleration is 3m/s/s straight forward, and the tension in the cable is 500N, solve for the drag force.

What is the lift provided by the parachute?

∑Fx = Fx + Fwind/Person = max

-500N * cos(45) + Fwind/Person = 70kg * -3m/s/s-353N + Fwind/Person = -210NFwind/Person = 243N

∑Fy = Fy + Fparachute/Person + Fg = 0-500N * sin(45) + Fparachute/Person + 70*9.8 = 0-353N + Fparachute/Person + 686N = 0 Fparachute/Person = 333N

Forces and Kinematics

The acceleration in F=ma is the same as the acceleration in x= at2+v0t+x0.

In these types of problems, you will either use forces to find acceleration and move to kinematics or start with kinematics to get the acceleration and move to forces.

Example

• A 50kg track star uniformly accelerates as he runs the first 25m in his race. Calculate the force necessary for him to run this in 4s.

Since we’re calculating force, we should start with kinematics.

a=?x=40mx0=0v0=0t=4s

x= at2+v0t+x0

40m = 0.5(a)(4s)^240m = 8aa = 5m/s/s

∑F = ma = (50kg)(5m/s/s) = 250N

Two-Body Problems

If two objects are connected so that they have the same acceleration, it can be useful to think of them as a system.

∑Fsystem = msys a

Example

A 2kg mass hangs off a table. A 6kg block is on the table. Assume friction is negligible. 1) Solve for the acceleration of the block and mass.2) Solve for the tension in the cable.

The force causing the block and mass to accelerate is the force of gravity on the hanging mass, so

∑Fsystem = Fg,mass= msys a(2kg)(-9.8N/kg) = (8kg)aa=-2.45m/s/s

To solve for the tension, look at either the individual mass or the individual block. I chose the block.

∑Fblock = mblock a = FCable/block

FCable/block = (2kg) (-2.45m/s/s)FCable/block = -4.9N

Alternatively, using the mass should yield the same result. Be careful of the signs. The mass is accelerating downward, which I have defined to be negative in this case.

∑Fmass = Fg+ FCable/mass =m a(2kg)(-9.8N/kg) + FCable/mass =(2kg)(-2.45m/s/s)

-19.6N + FCable/mass = -4.9NFCable/mass = 24.5N