A man with mass m1 = 59 kg stands at the left end of a uniform boat with mass m2 = 163 kg and a length L = 3.5 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water.
a. What is the location of the center of mass of the system?
b. If the man now walks to the right edge of the boat, what is the location of the center of mass of the system?
c. After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?)
d. After the man walks to the right edge of the boat, what is the new location the center of the boat?
e. Now the man walks to the very center of the boat. At what location does the man end up?
To answer these questions, we need to understand the concept of the center of mass.
The center of mass is the point where the entire mass of an object or system can be considered to be concentrated. In a uniform object or system, such as the boat-person system in this case, the center of mass can be found by taking the average of the individual masses weighted by their distances from a reference point.
Now let's go through each question step by step:
a. To find the location of the center of mass of the entire boat-person system at its original state, we need to use the formula for the center of mass of a system.
Let's assume that the boat is placed along the x-axis, with the left end of the boat at the origin (x = 0). Since the boat is uniform, the center of mass of the boat is at its geometrical center, which is L/2 = 3.5/2 = 1.75 m from the left end of the boat.
The center of mass of the entire system is given by the equation:
x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)
where x_cm is the location of the center of mass, m1 and m2 are the masses of the person and the boat respectively, and x1 and x2 are the positions of the person and the boat's center of mass respectively.
Plugging in the values, we have:
x_cm = (59 kg * 0 + 163 kg * 1.75 m) / (59 kg + 163 kg)
= 1.27 m (rounded to two decimal places)
So, the location of the center of mass of the system at its original state is at x = 1.27 m.
b. When the man walks to the right edge of the boat, we need to recalculate the center of mass. In this case, the position of the person changes, but the position of the boat's center of mass remains the same.
The new position of the person is at x = L = 3.5 m (the right edge of the boat).
Using the same formula as before, we get:
x_cm = (59 kg * 3.5 m + 163 kg * 1.75 m) / (59 kg + 163 kg)
= 2.52 m (rounded to two decimal places)
So, the location of the center of mass of the system when the man walks to the right edge of the boat is at x = 2.52 m.
c. To find how far the man has moved from his original location, we simply subtract the original position from the new position.
Distance moved by the man = New position - Original position
= 3.5 m - 0 m
= 3.5 m
So, the man has moved 3.5 m from his original location.
d. After the man walks to the right edge of the boat, the position of the boat's center of mass remains the same. Therefore, the new location of the center of mass is the same as in part b, i.e., x = 2.52 m.
e. Now, when the man walks to the very center of the boat, his new position is at x = L/2 = 3.5/2 = 1.75 m.
Using the same formula as before, we get:
x_cm = (59 kg * 1.75 m + 163 kg * 1.75 m) / (59 kg + 163 kg)
= 1.75 m
So, the man ends up at the very center of the boat, at x = 1.75 m.
I hope this helps! Let me know if you have any further questions.