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# Magnitude and direction of resultant forces

Asked Feb 28, 2009, 05:47 AM — 2 Answers
Sorry I forgot to put the values for the angl Q

All angls are measured anti clockwise from the x a-axis

F1=60N, F2=45N,F3=25N, F4=15N
Q1=30, Q2=120,Q3=220, Q4=320

2 Answers
 ebaines Posts: 10,039, Reputation: 5534 Expert #2 Mar 2, 2009, 01:24 PM
For each force vector break it into its x- and y-components. For example, the first force is F = 60N at an angle of 30 degrees, so it's x-component is 60*cos(30) and its y-component is 60*sin(30). Do this for all 4 forces. Then add up all the x-component forces to get a total force in the x-direction - let's call that Fx. Then do the same for the why direction to get Fy. The resultant total force is the sum of these two perpendicular force vectors: the magnitue of F is sqrt (Fx^2 + Fy^2), at an angle of Q= arctan(Fy/Fx). Post back and let us know what you get.
 kgshobana Posts: 4, Reputation: 1 New Member #3 Mar 14, 2009, 03:32 AM
Hi

you didn't mention whether the forces are tensile or compressive. If all the forces are tensile(i.e projecting outwards from the centre), the magnitude and the direction of resultant force for the given forces is 80.47 N and 63.25 deg respectively.
V all know the magnitude can be found out by taking square root of squares of the given forces and the direction by taking inverse tan of vertical to horizontal forces. For that resolve the given forces into vertical and horizontzl components and find out its cumulative by considering positive sign for towards right and upwards and negative for towards left and downwards.

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