A particle is located at the origin and is free to move into the x- y plane. Thr

Question

A particle is located at the origin and is free to move into the x- y plane. Three forces act on this particle. A force of 5.0 N, 40 degrees counter clockwise from the x axis, a force of 7.0 N, 120 degrees counter clockwise from the x axis and a force of 10.0 N, 230 degrees counter clockwise from the x axis.

a) Write all three forces in unit vector notation.
b) What is the magnitude and direction of a single additional force that would cause the particle to remain stationary? My Work: Ax = 5.0 cos(40) =-3.33 i N                
Ay = 5.0 sin (40) = 3.72 j N     
Bx = -7.0 cos(120) = -5.69 i N
By = 7.0 sin(120) = 5.21 j N
Cx = -10.0 cos(230) 7.87 i N
Cy = -10.0 sin(230) = 6.16 j N
Dx = Ax + Bx + Cx Dx = -3.33 + (-5.69) + 7.87 = -1.15 N
Dy = Ay + By + Cy Dy = 3.72 + 5.21 + 6.16 = 15.09 N
D = 1.15 i N + 15.09 j N b) |D| = v1.15^2 + 15.09^2 |D| = 15.13 N ? = tan -1 (15.09/1.15) = 13.12
Is this even correct?
b) What is the magnitude and direction of a single additional force that would cause the particle to remain stationary? My Work: Ax = 5.0 cos(40) =-3.33 i N                
Ay = 5.0 sin (40) = 3.72 j N     
Bx = -7.0 cos(120) = -5.69 i N
By = 7.0 sin(120) = 5.21 j N
Cx = -10.0 cos(230) 7.87 i N
Cy = -10.0 sin(230) = 6.16 j N
Dx = Ax + Bx + Cx Dx = -3.33 + (-5.69) + 7.87 = -1.15 N
Dy = Ay + By + Cy Dy = 3.72 + 5.21 + 6.16 = 15.09 N
D = 1.15 i N + 15.09 j N b) |D| = v1.15^2 + 15.09^2 |D| = 15.13 N ? = tan -1 (15.09/1.15) = 13.12
Is this even correct?
Cx = -10.0 cos(230) 7.87 i N
Cy = -10.0 sin(230) = 6.16 j N
Dx = Ax + Bx + Cx Dx = -3.33 + (-5.69) + 7.87 = -1.15 N
Dy = Ay + By + Cy Dy = 3.72 + 5.21 + 6.16 = 15.09 N
D = 1.15 i N + 15.09 j N b) |D| = v1.15^2 + 15.09^2 |D| = 15.13 N ? = tan -1 (15.09/1.15) = 13.12
Is this even correct?

Explanation / Answer

Ax = 5.0 cos(40) =-3.83 i N                
Ay = 5.0 sin (40) = 3.21 j N     
Bx = -7.0 cos(120) = 3.5 i N
By = 7.0 sin(120) = 6.06 j N
Cx = -10.0 cos(230) = 6.42 i N
Cy = -10.0 sin(230) = 7.66 j N
Dx = Ax + Bx + Cx Dx = 3.83 + (3.5) + 6.42 = 13.7 N
Dy = Ay + By + Cy Dy = 3.21+ 6.06 + 7.66 = 16.93 N
D = 13.7 i N + 16.93 j N b) |D| = v D_x ^2 +D_y ^2     |D| = v ( 13.7 )^2 + ( 16.93 )^2                       = 21.7 N ? = tan -1 (16.93/13.7) = 51 degrees
Ax = 5.0 cos(40) =-3.83 i N                
Ay = 5.0 sin (40) = 3.21 j N     
Bx = -7.0 cos(120) = 3.5 i N
By = 7.0 sin(120) = 6.06 j N
Cx = -10.0 cos(230) = 6.42 i N
Cy = -10.0 sin(230) = 7.66 j N
Dx = Ax + Bx + Cx Dx = 3.83 + (3.5) + 6.42 = 13.7 N
Dy = Ay + By + Cy Dy = 3.21+ 6.06 + 7.66 = 16.93 N
D = 13.7 i N + 16.93 j N b) |D| = v D_x ^2 +D_y ^2     |D| = v ( 13.7 )^2 + ( 16.93 )^2                       = 21.7 N ? = tan -1 (16.93/13.7) = 51 degrees
Cx = -10.0 cos(230) = 6.42 i N
Cy = -10.0 sin(230) = 7.66 j N
Dx = Ax + Bx + Cx Dx = 3.83 + (3.5) + 6.42 = 13.7 N
Dy = Ay + By + Cy Dy = 3.21+ 6.06 + 7.66 = 16.93 N
D = 13.7 i N + 16.93 j N b) |D| = v D_x ^2 +D_y ^2     |D| = v ( 13.7 )^2 + ( 16.93 )^2                       = 21.7 N ? = tan -1 (16.93/13.7) = 51 degrees

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