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# Concurrent forces 3p,7p and 5p act respectively along three directions which are parallel to side of an equilateral triangle taken in order determine the magnitude and direction of the resultant

abhay gupta
4 months ago
Dear student,

For an equilateral triangle, all angles are equal A=B=C=60 degree

Sum of forces acting in HORIZONTAL direction:

F1 = 3P + 5P (-cos60) + 7P (-cos60)

= 3P + 5P (-1/2) + 7P (-1/2)

= -6P

Sum of forces acting in VERTICAL direction:

F2 = 0 + 5P (Sin60) + 7P (-sin60)

= 5P (0.866) + 7P (-0.866)

= -1.732P

Magnitude of resultant force (R)= Square root (F1 + F2)

= square root ((-6)^2 + (-1.732)^2))

= square root (39)

MagnitudeR = 6.24P

Let ϴ be the angle of the resultant force in direction x

Then ϴ = tan-1 [(-6P)/(-1.732P)]

Direction of resultant force:

ϴ = 73.89 degree

Ankit
Vibekjyoti Sahoo
145 Points
17 days ago
For an equilateral triangle, all angles are equal A=B=C=60 degree

Sum of forces acting in HORIZONTAL direction:

F1 = 3P + 5P (-cos60) + 7P (-cos60)

= 3P + 5P (-1/2) + 7P (-1/2)

= -6P

Sum of forces acting in VERTICAL direction:

F2 = 0 + 5P (Sin60) + 7P (-sin60)

= 5P (0.866) + 7P (-0.866)

= -1.732P

Magnitude of resultant force (R)= Square root (F1 + F2)

= square root ((-6)^2 + (-1.732)^2))

= square root (39)

MagnitudeR = 6.24P

Let ϴ be the angle of the resultant force in direction x

Then ϴ = tan-1 [(-6P)/(-1.732P)]

Direction of resultant force:

ϴ = 73.89 degree