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Ship at Sea, Apparent Weight Problem

by physics114
Tags: apparent, crewman, ship, weight
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physics114
#1
Oct21-10, 02:17 AM
P: 7
Correct? Please let me know if my logic is off. Thank you =]

Given Problem
At the bow of a ship on a stormy sea, a crewman conducts an experiment by standing on a bathroom scale. In calm waters, the scale reads 180 lb. During the storm, the crewman finds a maximum reading of 234 lb and a minimum reading of 136 lb
(a)Find the maximum upward acceleration experienced by the crewman. (m/s2?)
(b)Find the maximum downward acceleration experienced by the crewman. (m/s2?)

Relevant Equations
W = mg
F = ma
Wa = m (g + a) (Acceleration Upwards)
Wa = m (g - a) (Acceleration Downwards)
1 lb = 4.45N

Attempted Solution
So I was trying to figure out if there was somethin' special about the calm water and stormy water but I am still lost... Does the fact that the 180lb is only read during calm water mean that at that moment it is the shipman's true weight because there is not upward or downward acceleration acting on it?

If the first assumption about the calm vs. stormy water is correct than I think I would go at it by using the W=mg equation to find the shipman's mass.. using the 180lb (convert to N?) as mass and 9.8m/s2 for gravity.

Find mass of suitcase:
180lb = 801N
W = mg
m = W / g
m = 801N / 9.8m/s2
m = 81.7347kg

(a) Find Max Upward Acceleration Experienced by Crewman
Using 234lb as the highest possible number
234lb = 1041.3N
Use formula for upwards acceleration
Wa = mg + ma
Wa - mg = ma
a = (Wa - mg) / m
a = [(1041.3N) - (81.7347kg * -9.8m/s2)] / (81.7347kg)
acceleration is -9.8m/s2 instead of 9.8m/s2 because acceleration is upward?
a = [(1041.3) - (-801)] / (81.7347)
a = 1842.3 / 81.7347
a = 22.54 m/s2

(b) Find Max Downward Acceleration Experienced by Crewman
Follow the same steps as in part a but use the minimum of 136lb as the apparent weight instead (?)
136lb = 605.2N
Using formula for downward acceleration (Wa = mg - ma)
a = Wa + mg/ m
a = 605.2 + (-801) / 81.7347
a = 195.8 / 81.7347
a = -2.39556 m/s2

Yay?
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Borek
#2
Oct21-10, 03:15 AM
Admin
Borek's Avatar
P: 23,716
Without any analysis it looks like you are wrong. Up and down accelerations you have listed differ by the order of magnitude, while changes in weight (+54,-44) are very similar.


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