Two forces F_1 and F_2 act on a 1.60-kg object. F_1 = 35.0 N and F_2 = 16.0 N. F

Question

Two forces F_1 and F_2 act on a 1.60-kg object. F_1 = 35.0 N and F_2 = 16.0 N. Find the acceleration of the object for the configuration of forces shown in Figure (a), Find the acceleration of the object for the configuration of forces shown in Figure (b). magnitude You find the magnitude of a force in the same manner you find the magnitude of any other vector: first resolve each vector into components, m/s^2 You can determine the angle from the components of the net force, or from the components of the acceleration, (counterclockwise from F_1)

Explanation / Answer

b) Fx = F1 + F2*cos(theta)

= 35 + 16*cos(60)

= 43 N

Fy = F2*sin(60)

= 16*sin(60)

= 13.86 N

|F| = sqrt(Fx^2 + Fy^2)

= sqrt(43^2 + 13.86^2)

= 45.2 N

|a| = |F|/m

= 45.2/1.6

= 28.2 m/s^2 <<<<<<<<-----------Answer

direction : theta = tan^-1(Fy/Fx)

= tan^-1(13.86/43)

= 17.9 degrees with From F1 <<<<<<<<-----------Answer

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