A- What is the magnitude of the resultant of these two forces? Answer in units o

Question

A- What is the magnitude of the resultant of these two forces? Answer in units of N.

B- Find the direction of the resultant force (in relation to forward, with counterclockwise considered positive). Answer in degrees from the positive x-axis, with counter-clockwise positive, within the limits of 180 to 180 . Answer in units of .

C-If the car has a mass of 3186 kg, what acceleration does it have? Ignore friction. Answer in units of m/s2.

Two forces, 383 N at 12 degree and 339 N at 29 degree are applied to a car in an effort to accelerate it. A- What is the magnitude of the resultant of these two forces? Answer in units of N. B- Find the direction of the resultant force (in relation to forward, with counterclockwise considered positive). Answer in degrees from the positive x-axis, with counter-clockwise positive, within the limits of -180 degree to 180 degree. Answer in units of degree. C-If the car has a mass of 3186 kg, what acceleration does it have? Ignore friction. Answer in units of m/s2.

Explanation / Answer

forces F = 383 N           F ' = 339 N In vertical direction : -------------------- net force f = 339 sin 29 - 383 sin 12                  = 84.72 N     downward In horizontal direction : ---------------------- net force f ' = 383 cos12 + 339 cos29                   = 671.126 N   in forward direction resultant force vector f " = 671.126 i -84.72 j magnitude of f " = [ 671.126 ^ 2 + (-84.72 ) ^ 2]                          = 676.45 N (b). let f " makes an angle with horizontal along horizontal direction then tan = f / f '            = 0.126        = 7.194 degrees   it is clock wise direction to horizontal So, = -7.194 degrees (c). mass m = 3186 kg accleration a = f " / m                     = 0.2123 m / s^ 2

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