Homework Solutions for
Dynamics
Homework 8 Solutions
4. Without friction, the only horizontal force is
the tension. We apply Newton’s second law to
the car:
F = ma;
FT = (1050 kg)(1.20 m/s2) = 1.26 x 103 N.
18. With down positive, we write ·F = ma from the force
diagram for the skydivers:
mg – FR = ma;
(a) Before the parachute opens, we have
mg – .25mg = ma, which gives a = +g = 7.4 m/s2 (down).
(b) Falling at constant speed means the acceleration is
zero, so we have mg – FR = ma = 0, which gives
FR = mg = (120.0 kg)(9.80 m/s2) = 1176 N.
Do Now
The cable supporting a 2100 kg elevator has
a maximum strength of 21,750 N. What
maximum upward acceleration can it give
the elevator without breaking? Begin with a
free body diagram.
Homework 9 Solutions
20. We find the velocity necessary for the jump from the motion when the
person leaves the ground to the highest point, where the velocity is zero:
v2 = vjump2 + 2(– g)h;
0 = vjump2 + 2(– 9.80 m/s2)(0.80 m), which gives vjump = 3.96 m/s.
We can find the acceleration required to achieve this velocity during the crouch
from vjump2 = v02 + 2a(y – y0);
(3.96 m/s)2 = 0 + 2a(0.20 m – 0), which gives a = 39.2 m/s2.
Using the force diagram for the person during the crouch, we can write ·F = ma:
FN – mg = ma;
FN – (66 kg)(9.80 m/s2) = (66 kg)(39.2 m/s2), which gives FN = 3.2 x 103 N.
From Newton’s third law, the person will exert an equal and opposite force on the
Ground: 3.2 x 103 N downward.
21. (a) We find the velocity just before striking the ground from
v12 = v02 + 2(– g)h;
v12 = 0 + 2(9.80 m/s2)(4.5 m), which gives v1 = 9.4 m/s.
(b) We can find the average acceleration required to bring the person to
rest from v2 = v12 + 2a(y – y0);
0 = (9.4 m/s)2 + 2a(0.70 m – 0), which gives a = – 63 m/s2.
Using the force diagram for the person during the crouch,
we can write ·F = ma: mg – Flegs = ma;
(45 kg)(9.80 m/s2) – Flegs = (45 kg)(– 63 m/s2), which gives Flegs = 3.3 x 103 N up
Do Now
A 75 kg thief wants to escape from a 3rd story jail window.
Unfortunately, a makeshift rope made of sheets tied
together can support a mass of only 58 kg. How might the
thief use this ‘rope’ to escape? Give quantitative
evidence for your answer. Start with a FBD.
Homework 11 Solutions
Note: m1g in the lower diagram should read m2g
32. (a) Because the speed is constant, the acceleration
is zero.
We write ·F = ma from the force diagram:
FT + FT – mg = ma = 0, which gives
2FT = mg = 1/2(65 kg)(9.80 m/s2) = 3.2 x 102 N.
(b) Now we have:
FT’ + FT’ – mg = ma;
2(1.10)(mg) – mg = ma, which gives
a = 0.10g = 0.10(9.80 m/s2) = 0.98 m/s2
Do Now
An elevator in a tall building is allowed to reach a maximum
speed of 3.5 m/s going down. What must the tension be in the
cable to stop this elevator over a distance of 3.0 m if the
elevator has a mass of 1300 kg including its occupants?
Homework 12 Solutions
Do Now:
On an icy day,, you worry about parking your car in
your driveway, which has an incline of 12o. Your
neighbor Ralph’s driveway has an incline of 9.0o and
Bonnie’s driveway across the street has one of 6.0o.
The coefficient of static friction between tire rubber
and ice is 0.15. Which driveway(s) will be safe to park
in?
Do Now
A small block of mass m is given an initial speed v0 up a
ramp inclined at angle θ to the horizontal. It travels a
distance d up the ramp and comes to rest.
a) Determine a formula for the coefficient of kinetic
friction between the block and ramp.