Question

1. A mass of m = 6 kg has been accelerated by three separate forces acting in different directions in 2-dimensional space. Let each force be given by its acceleration vector in polar coordinates (R, q):

a₁ = 10 m/s² at an angle of 45^o

a₂ = 15 m/s² at an angle of 130^o

a₃ = 4 m/s² at an angle of 320^o

1. Find the magnitude of the resultant acceleration and the direction at which it is pointing.
2. What is the resultant magnitude of the force on the mass m?

Answer 1

Net acceleration on the mass will be:

a_net = a1+ a2 + a3

a1 = 10 m/s^2 at 45 deg

a2 = 15 m/s^2 at 130 deg

a3 = 4 m/s^2 at 320 deg

then, we know that, Suppose given that Vector is R and it makes angle A with +x-axis, then it's components are given by:

Rx = R*cos A

Ry = R*sin A

Using above rule:

a_net = ax_net + ay_net

a_net = (a1x + a2x + a3x) i + (a1y + a2y + a3y) j

Using above values

a_net = (10*cos 45 deg + 15*cos 130 deg + 4*cos 320 deg) i + (10*sin 45 deg + 15*sin 130 deg + 4*sin 320 deg) i

a_net = (7.07 - 9.64 + 3.06) i + (7.07 + 11.49 - 2.57) j

a_net = 0.49 i + 15.99 j

Magnitude of net acceleration will be:

|a_net| = magnitude of net acceleration = sqrt (ax_net^2 + ay_net^2)

|a_net| = sqrt ((0.49)^2 + (15.99)^2)

|Fnet| = 16.0 N

Direction of net acceleration will be

Direction = arctan (ay_net/ax_net)

Direction = arctan (15.99/0.49) = 88.24 deg above +ve x-axis

Part B.

Using Newton's 2nd law:

F_net = m*a

m = mass of object = 6 kg

F_net = 6*16.0

F_net = 96.0 N

Let me know if you've any query.