**Especially part a)
A person, sunbathing on a warm day, is lying horizontally on the deck of a boat. Her mass is 54.6 kg, and the coefficient of static friction between the deck and her is 0.683. Assume that she is moving horizontally, and that the static frictional force is the only force acting on her in this direction. (a) What is the magnitude of the static frictional force when the boat moves with a constant velocity of +5.95 m/s? (b) The boat speeds up with an acceleration of 1.59 m/s2, and she does not slip with respect to the deck. What is the magnitude of the static frictional force that acts on her? (c) What is the magnitude of the maximum acceleration the boat can have before she begins to slip relative to the deck?
THANKS
a. constant velocity means no acceleration.
F=ma=0
b. F=ma
Now, c( maximum acceleration
mu*mg=ma
a=mu*g
To solve these problems, we need to use the concept of static friction. It is the force that keeps the person from sliding or slipping on the deck. The static frictional force can adjust its magnitude up to a certain limit to prevent motion.
(a) When the boat moves with a constant velocity of +5.95 m/s, it means there is no acceleration. Therefore, the net force acting on the person is zero.
To find the magnitude of the static frictional force, we need to consider the equation for static friction:
fs = μs * N
Where fs is the static frictional force, μs is the coefficient of static friction, and N is the normal force.
The normal force N is equal in magnitude and opposite in direction to the gravitational force acting on the person, given by:
N = mg
Where m is the mass of the person and g is the acceleration due to gravity.
Now, substituting the values:
m = 54.6 kg,
μs = 0.683,
g = 9.8 m/s².
N = (54.6 kg) * (9.8 m/s²) = 535.08 N.
Finally, calculating the static frictional force:
fs = μs * N = (0.683) * (535.08 N) = 365.76 N.
So, the magnitude of the static frictional force when the boat moves with a constant velocity of +5.95 m/s is 365.76 N.
(b) In this case, the boat speeds up with an acceleration of 1.59 m/s², and the person does not slip with respect to the deck. We need to find the new magnitude of the static frictional force.
The net force acting on the person is given by:
ΣF = fs - m * a
Where a is the acceleration of the person.
Since the person does not slip with respect to the deck, the static frictional force fs is equal to the force required to provide the necessary acceleration:
fs = m * a
Now, substituting the values:
m = 54.6 kg,
a = 1.59 m/s².
The static frictional force is:
fs = (54.6 kg) * (1.59 m/s²) = 86.694 N.
So, the magnitude of the static frictional force that acts on her when the boat speeds up with an acceleration of 1.59 m/s² is 86.694 N.
(c) To find the maximum acceleration the boat can have before the person begins to slip relative to the deck, the static frictional force must reach its maximum value.
The maximum static frictional force is given by:
fmax = μs * N
Now substituting the values:
μs = 0.683,
N = (54.6 kg) * (9.8 m/s²) = 535.08 N.
fmax = (0.683) * (535.08 N) = 365.76 N.
The maximum acceleration can be found using:
fmax = m * a
Rearranging the equation:
a = fmax / m
a = (365.76 N) / (54.6 kg) = 6.70 m/s².
So, the magnitude of the maximum acceleration the boat can have before the person begins to slip relative to the deck is 6.70 m/s².