Grade 12th passMechanics

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

Vijaykumar

5 Years agoGrade 12th pass

2 Answers

abhay gupta

5 Years ago

Dear student,
Here’s your answer

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

Askiitians expert
Ankit

Vibekjyoti Sahoo

4 Years 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