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 =ma
FN + (50kg * -9.8N/kg) = 50kg * 2 m/s2
FN – 490N = 100N
FN = 590N
Example 2
A 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 = -210N
Fwind/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
1 2
acceleration in x= at +v0t+x0.
2
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.
1
x=
2
at2+v0t+x0
40m = 0.5(a)(4s)^2
40m = 8a
a = 5m/s/s
∑F = ma
= (50kg)(5m/s/s)
= 250N
a=?
x=40m
x0=0
v0=0
t=4s
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)a
a=-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 = -4.9N
FCable/mass = 24.5N