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 Oct 21st 2013, 10:15 AM #1 Junior Member   Join Date: Oct 2013 Posts: 6 Hi guys I have 2 questions and I need your help please ^__^ 1: Using components determine the sum of + if = 3 m [N30°W] and = 2 m [N10°E]. 2: A force of 3.0 N and 1.0 N act as shown. What is the acceleration of the 6.0 kg mass?
Oct 21st 2013, 01:05 PM   #2

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 Originally Posted by Duha 1: Using components determine the sum of + if = 3 m [N30°W] and = 2 m [N10°E].
Please let me be the first to say: "What??" This makes no sense. Can you write out the full problem?

 Originally Posted by Duha 2: A force of 3.0 N and 1.0 N act as shown. What is the acceleration of the 6.0 kg mass?
For this one you'll want to find the net force acting on the block in components in the usual +x and +y directions. Do you know how to do this?

-Dan
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 Oct 21st 2013, 01:41 PM #3 Junior Member   Join Date: Oct 2013 Posts: 6 Hi Im so sorry with the first question 1) 1: Using components determine the sum of a + b if a = 3 m [N30°W] and b = 2 m [N10°E]. And the second question, by using sine and cosine for horizontal and vertical but i dont know how to sort the solution. Thanks a lot for you.
Oct 21st 2013, 03:40 PM   #4

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 Originally Posted by Duha 1) 1: Using components determine the sum of a + b if a = 3 m [N30°W] and b = 2 m [N10°E].
Let's start simple:
Call East the +x axis and North the +y axis. Then we know that a is 3 m long at an angle of 30 degrees above the -x axis. So if we draw the usual triangles to get the components we have that a_x = - (3 m)*cos(30) and a_y = (3 m)*sin(30).

What are the components of the b vector?

-Dan
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 Oct 24th 2013, 09:36 AM #5 Junior Member   Join Date: Oct 2013 Posts: 6 Im so sorry for being late The components of vector b it will be (2 m)*cos(10) and (2 m)* sin(10) ?!
Oct 24th 2013, 12:29 PM   #6
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 Originally Posted by Duha Im so sorry for being late The components of vector b it will be (2 m)*cos(10) and (2 m)* sin(10) ?!
Yes, correct.

 Oct 24th 2013, 12:39 PM #7 Junior Member   Join Date: Oct 2013 Posts: 6 And then a + b will be ????
 Oct 24th 2013, 01:57 PM #8 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 Add a_x and b_x to get the total x-component, then add a_y and b_y to get the total y-component. The magnitide of the resulting vector is from Pythagoras: sqrt(x^2 + y^2). Its direction is calculated using trig: arctan(angle) = x/y to get angle east or west from north.
 Oct 26th 2013, 10:27 AM #9 Junior Member   Join Date: Oct 2013 Posts: 6 So for a_x will be (3 m)*cos(30) + (2 m)*cos(10) and a_y (3 m)*sin(30) + (2 m)*sin(10) is that right ????!!!!
 Oct 30th 2013, 10:27 PM #10 Junior Member   Join Date: Oct 2013 Posts: 6 I solved the first question guys Could you please now help me with the second one ^__^

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