2 Forces problems (Halliday/Resnick/Walker, 7th Ed. book)

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The discussion focuses on solving two physics problems from Halliday/Resnick/Walker, 7th Ed. The first problem involves calculating the third force acting on a lemon half on a frictionless table, given two horizontal forces, with conditions for stationary, constant, and varying velocities. The second problem addresses the tension in a cable pulling an elevator cab, where a dropped coin accelerates downward at 8.00 m/s². The calculations for both problems utilize Newton's second law, with the first problem yielding specific force values and the second determining the tension as 3600 N. Participants express a need for clarification on the first problem's visualization and the application of Galilean relativity.
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Homework Statement


In chapter 5, problems 26:
A provided figure shows an overhead view of a 0.0250 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F_1 has a magnitude of 6.00 N and is at θ_1 = 30.0° (and, btw, is in quadrant II). Force F_2 has a magnitude of 7.00 N and is at θ_2 = 30.0° (in quadrant IV). In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has constant velocity v= (13.0 i- 14.0 j) m/s, and (c) has varying velocity v= (13.0t i- 14.0t j) m/s^2, where t is time?


Homework Equations


F_net = m*a
F_net = F_1 + F_2 + F_3
F_3 = ?
F_3 = F_net- F_1 - F_2
needed: F_3x = m*a_x - F_1*cosθ_1 - F_2sinθ_2
F_3y = m*a_y - F_1*sinθ_1 - F_2cosθ_2



The Attempt at a Solution


(a) acceleration = 0, therefore:
F_3 = 9.00 N i- 9.06 N j

(b) acceleration is still zero. so i assume part (a) = part (b)

(c)
F_net= m*a
m = 0.0250 kg
a = (sqrt) [13.0^2 + (-14.0)^2] = 19.1 m/s^2
F_net=0.4776 N
F_3= -8.37 N i- 9.41 N j


Homework Statement


(Second problem) Chapter 5, problem 36:
An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 2000 kg. When that occupant drops a coin, its acceleration relative to the cab is 8.00 m/s^2 downward. What is the tension in the cable?


Homework Equations


I used:
T= mg-ma


The Attempt at a Solution


T= 2000 kg (9.80 m/s^2)- 2000 kg (8.00 m/s^2)
= 3600 N

---please HELP! :eek: :bugeye: :cry:
 
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I can't really visualize your first problem from your description, but from what I picked up I think you did it right.

Did you derive your tension equation from Newton's second law? What does galilean relativity say about acceleration?
 
Thank you Mindscrape.

Yes the tension equation is from Newton's second law. I had to research the Galilean relativity...ermmm. I'll just ask my professor later. Thank you though!
 
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