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Mathematics: Post your doubts here!

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Nov012 v.42 no4 :S
What they have given us:The 'x' components of the following forces:
68 N = -60N
75N= 0N
100N= 96N,
What we gotta find, i) (resolve the forces 100 N and 68N) therefore, 68cos(theta) = -60 , theta= 28.1 deg
100cos (beta) =96, beta=16.2 deg, hence we now know the angles that the forces make with the horizontal axis. We can find 'y' components of the forces.
68 N => -68sin28.1 = -32 N ( negative as it in he the negative y axis direction)
75 N => no components to resolve as it is itself on the y-axis so, 75 N
100 N => -100sin16.2= -27.9 N

ii) List the Fx components of the forces And then the Fy components as we calculated above:
Fx=> -60+0+96 = 36 N
Fy=> -32+75-27.9= 15.1 N
Resultant = (36^2 + 15.1^2)^1/2 (i took the square root of the squares of the sum of the two forces we calculated (Fx and Fy)
Therefore, tan(theta)= y/x => tan (theta) = 15.1/36, theta= 22.8 deg.
Hope u got it. :D
 
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Math_zpsa852621e.png
What they have given us:The 'x' components of the following forces:
68 N = -60N
75N= 0N
100N= 96N,
What we gotta find, i) (resolve the forces 100 N and 68N) therefore, 68cos(theta) = -60 , theta= 28.1 deg
100cos (beta) =96, beta=16.2 deg, hence we now know the angles that the forces make with the horizontal axis. We can find 'y' components of the forces.
68 N => -68sin28.1 = -32 N ( negative as it in he the negative y axis direction)
75 N => no components to resolve as it is itself on the y-axis so, 75 N
100 N => -100sin16.2= -27.9 N

ii) List the Fx components of the forces And then the Fy components as we calculated above:
Fx=> -60+0+96 = 36 N
Fy=> -32+75-27.9= 15.1 N
Resultant = (36^2 + 15.1^2)^1/2 (i took the square root of the squares of the sum of the two forces we calculated (Fx and Fy)
Therefore, tan(theta)= y/x => tan (theta) = 15.1/36, theta= 22.8 deg.
Hope u got it. :D



Thaaaanks a millioooooon , May Allaaah bless youu both :)
 
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