1) An object acted on by three forces moves with constant velocity. One force ac

Question

1) An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.4 N ; a second force has a magnitude of 4.6 N and points in the negative y direction.

Find the direction of the third force acting on the object.

2) An object of mass 6.80 kg has an acceleration a =(1.26 m/s2 )x^+(-0.699 m/s2 )y^.

Three forces act on this object: F1, F2,and F3.Given that F1=(3.49 N )x^ and F2=(-1.56 N )x^+(1.82 N )y^, find F3.

F3x, F3y = _______________N

Explanation / Answer

Given,

F1 = 6.4 i N

F2 = - 4.6 j N

Third force acting on the body to keep it to move with uniform speed is

F3 = - (F1 + F2) = - (6.4 i - 4.6 j) = -6.4 i + 4.6 j N

Magnitude = |F3| = Sqrt[(-6.4)^2 + (4.6)^2] = 7.88 N

Direction = theta = arctan [4.6/-6.4] = - 35.71degrees

So, direction, theta = -35.71 + 180 = 144.29 degree counter clock wise from +ve x axis.

Comment in case any doubt please rate my answer....

According to Chegg guidelines I can answer one question at a time so please post other questions separately....

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.