Three forces in the x-y plane act on a 6.80 kg mass: 10.10 N directed at 19o, 8.60 N directed at 122o, and 9.80 N directed at 218o. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction. Calculate the magnitude of the acceleration.

Answer: a = 1.58 m/s²

Explanation: The mass of the object is 6.80kg.

From newton's second law of motion,

Resultant force = mass * acceleration

So we need to get the resultant force first.

Note all angles are measured from the positive x axis in the counter clockwise direction ...

The first force is 10.10 N at an angle of 19° to the positive x axis.

Horizontal component of this force = 19 * cos 19° = 17.964 N

Vertical component of this force = 19 * sin 19° = 6.186N

The second force is 8.6 N, 122° (22° to the positive y axis), due to the angle, the force is between the positive y axis and negative x axis (2nd quadrant)

Horizontal component of this force = - 8.6*sin 22 = - 3.22 N

Vertical component of this force = 8.6 * cos 22 = 7.974 N

The third force is 9.8 N, 218° (38 degree to the negative axis), due to the angle, the force is placed in between the negative x and negative y axis.

Horizontal component of this force = - 9.8*cos 38 = - 7.722 N

Vertical component of this force = - 9.8 * sin 38 = - 6.033 N

Sum of forces on the x axis (fx) = 17.964 - 3.22 - 7.722 = 7.022 N

Sum of forces on the y axis (fy) = 6.186 + 7.974 - 6.033 = 8.127 N

Resultant force = √ (fx) ² + (fy) ²

Resultant force = √7.022² + 8.127²

Resultant force = √115.356613

Resultant force = 10.74 N

Recall that resultant force = mass * acceleration

10.74 = 6.8 * a

a = 10.74/6.8

a = 1.58 m/s²