Five forces act on an object: (1) 60.0 N at 90°, (2) 40 N at 0°, (3) 80 N at 270°, (4) 40 N at 180°, (5) 50 N?
Five forces act on an object: (1) 60.0 N at 90°, (2) 40 N at 0°, (3) 80 N at 270°, (4) 40 N at 180°, (5) 50 N at 60°. What are the magnitude and direction of a sixth force that would produce equilibrium?
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see the sum of all forces ( vector sum) must be zero.
the resultant force of above forces is 25 i + 25*(root of 3) - 20 j.
hence the vector which is equal in mag and opp in direcn will be the ans so that the total sum is 0
therefore the ans is -25 i and 20 - 25*(root of 3) j
where i is the unit vec along 0° and j is the unit vec along 90°.
58cos(ninety), 58sin(ninety) 40cos(0), 40sin(0) 85cos(270), 85sin(270) 40cos(a hundred and eighty), 40sin(a hundred and eighty) 50cos(60), 50sin(60) Rx = 40 - 40 + 25 = 25 Ry = fifty 8 - 80 5 + 25sqrt(3) = sixteen.3 The stress could could be - 25 interior the x direction and - sixteen.3 interior the y direction Rf = Sqrt(Rx^2 + Ry^2) = 29.845 N attitude = arctan(Ry/Rx) = 33.106 deg