Download dynamics problem solving – 2

KEY: DYNAMICS PROBLEM SOLVING – 2
Problems where all forces are parallel with coordinate axes
Instructions: Solve each of the following problems. Show all steps, including a sketch, motion diagrams,
free body diagrams, Newton’s second law applied to both the x and the y-axes, and the complete solution,
with units. Do not take shortcuts or neglect any work, or you will not get full credit for this assignment.
1. A cable pulls a 250-kg cart. If the cart’s speed increases at a rate of 3.2 m/s2, and the friction force is
50-N, what is the tension in the cable? (850 N)
Fx = max
T – Ff = max
T – 50 = 250 (3.2)
T- 50 = 800
T = 850 N
2. A 60 kg water skier is pulled at constant speed. If friction is 150-N, what is the tension in the rope? (150
N)
Fx = max
T – Ff = max
T – 150 = 60 (0)
T – 150 = 0
T = 150 N
3. A 60 kg bicyclist coasts along a road, decreasing speed at a rate of 7.8 m/s2. What is the magnitude of
the friction force? (468 N)
Fx = max
– Ff = max
– Ff = (60) (-7.8)
– Ff = -468
Ff = 468 N
4.
A 100-kg cart, initially at rest, is dragged along a frictionless horizontal surface with an acceleration of
7.8 m/s2. What is the tension?
Fx = max
T – Ff = max
T = 100 (7.8)
T = 780 N
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5. After rolling down an incline, a 500-kg cart moves along the frictionless horizontal surface until it is
stopped by plowing into a haystack. If the average force of the haystack on the cart is 3000-N, then
what is the acceleration of the cart while stopping?
Fx = max
– Nhay = max
– 3000 = (500) (a)
- 6 m/s2 = a
6. When the click beetle jumps in the air, its acceleration upward can be as large as 400.0 times the
acceleration due to gravity. (An acceleration this large would instantly kill any human being.) For a
beetle whose mass is 40.00 mg, calculate the magnitude of the force exerted by the beetle on the ground
at the beginning of the jump with gravity taken into account. Calculate the magnitude of the force with
gravity neglected. Use 9.807 m/s2 as the value for free-fall acceleration.
a = 400*g = 400 * 9.8 = 3923 m/s2
mbeetle = 40 milligrams = 0.00004 kg
Fy = may
N – w = may
N - (4 x 10-5)(9.8) = (4 x 10-5)(3923)
N – 3.923 x 10-4 = 0.1570
N = 0.1573 N
7. In 1994, a Bulgarian athlete named Minchev lifted a mass of 157.5 kg. By comparison, his own mass
was only 54.0 kg. Calculate the force acting on each of his feet at the moment he was lifting the mass
with an upward acceleration of 1.00 m/s2. Assume that the downward force on each foot is the same.
(HINT: You’ll need to draw FBDs for the two systems separately.)
For the barbell:
Fy = may
N – w = may
N - (157.5)(9.8) = (157.5)(1)
N – 1543.5 = 157.5
N = 1701 N
For the man:
Fy = may
Nfloor – w – Nbarbell = may
N - (54)(9.8) – 1701 = (54)(0)
N – 2230 = 0
Nfloor = 2230 N
Nfloor on each foot = Nfloor÷2 = 1115 N
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