Two forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 400 N, is applied at an angle α = 32° to the forward dashed line. The second force, F2 = 536 N, is applied at an angle β = 10° to the forward dashed line.

(a) What is the resultant of these two forces?
N at ° to the

of the forward dashed line

(b) If the car has a mass of 2900 kg, what acceleration does it have? (Disregard friction.)
m/s2 at ° to the

of the forward dashed line

F1 = 400N[32o]

F2 = 536N.[10o]

a. X = 400*Cos32 + 536*Cos10 =
Y = 400*sin32 + 536*sin10 =

Tan A = Y/X
A =

Fr = X/Cos A = Resultant force @ Ao.

b. a = Fr/m = Acceleration @ A Degrees.

To find the resultant of the two forces, we can use vector addition. We'll break down each force into its horizontal and vertical components, and then add up all the components separately.

(a) Resultant of the forces:
First, let's find the horizontal and vertical components of each force.

The horizontal component of F1 can be found using the formula:
F1_horizontal = F1 * cos(α)
F1_horizontal = 400 N * cos(32°)
F1_horizontal ≈ 340.97 N (rounded to two decimal places)

The vertical component of F1 can be found using the formula:
F1_vertical = F1 * sin(α)
F1_vertical = 400 N * sin(32°)
F1_vertical ≈ 209.14 N (rounded to two decimal places)

Similarly, we can find the horizontal and vertical components of F2.

The horizontal component of F2 can be found using the formula:
F2_horizontal = F2 * cos(β)
F2_horizontal = 536 N * cos(10°)
F2_horizontal ≈ 530.01 N (rounded to two decimal places)

The vertical component of F2 can be found using the formula:
F2_vertical = F2 * sin(β)
F2_vertical = 536 N * sin(10°)
F2_vertical ≈ 92.98 N (rounded to two decimal places)

Now, let's add up the horizontal and vertical components of the two forces separately.

Horizontal component of resultant = F1_horizontal + F2_horizontal
Horizontal component of resultant = 340.97 N + 530.01 N
Horizontal component of resultant ≈ 870.98 N (rounded to two decimal places)

Vertical component of resultant = F1_vertical + F2_vertical
Vertical component of resultant = 209.14 N + 92.98 N
Vertical component of resultant ≈ 302.12 N (rounded to two decimal places)

Using these horizontal and vertical components, we can find the magnitude and direction of the resultant force.

The magnitude of the resultant force can be found using the Pythagorean theorem:
Magnitude of resultant force = √(horizontal component^2 + vertical component^2)
Magnitude of resultant force = √(870.98^2 + 302.12^2)
Magnitude of resultant force ≈ 933.06 N (rounded to two decimal places)

The angle of the resultant force can be found using the inverse tangent function:
Angle of resultant force = tan^(-1)(vertical component / horizontal component)
Angle of resultant force = tan^(-1)(302.12 N / 870.98 N)
Angle of resultant force ≈ 19.549° (rounded to three decimal places)

Therefore, the resultant of the two forces is approximately 933.06 N at an angle of 19.549° to the forward dashed line.

(b) Now, let's calculate the acceleration of the car using Newton's second law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.

Acceleration = Force / Mass
Acceleration = Resultant force / Mass
Acceleration = 933.06 N / 2900 kg
Acceleration ≈ 0.322 m/s² (rounded to three decimal places)

Additionally, the angle of the acceleration can be determined by using the angle of the resultant force. In this case, the angle of the acceleration will be the same as the angle of the resultant force.

Therefore, the car has an acceleration of approximately 0.322 m/s² at an angle of 19.549° to the forward dashed line.