Two forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 328 N, is applied at an angle α = 35° to the forward dashed line. The second force, F2 = 464 N, is applied at an angle β = 13° to the forward dashed line.
(a) What is the resultant of these two forces?
(b) If the car has a mass of 2950 kg, what acceleration does it have? (Disregard friction.)
To solve this problem, we need to find the resultant force (a) and then use that to calculate the acceleration of the car (b).
(a) To find the resultant force, we can use vector addition. We can break down each force into its x and y components.
The x-component of F1 is given by F1x = F1 * cos(α).
F1x = 328 N * cos(35°) = 268.76 N.
The y-component of F1 is given by F1y = F1 * sin(α).
F1y = 328 N * sin(35°) = 186.34 N.
Similarly, the x-component of F2 is given by F2x = F2 * cos(β).
F2x = 464 N * cos(13°) = 448.95 N.
The y-component of F2 is given by F2y = F2 * sin(β).
F2y = 464 N * sin(13°) = 107.45 N.
Next, we can add the x-components and y-components separately to get the resultant force vector.
Sum of x-components = F1x + F2x = 268.76 N + 448.95 N = 717.71 N.
Sum of y-components = F1y + F2y = 186.34 N + 107.45 N = 293.79 N.
Using these values, we can find the magnitude and direction of the resultant force.
Magnitude of the resultant force, R = sqrt((Sum of x-components)^2 + (Sum of y-components)^2).
R = sqrt((717.71 N)^2 + (293.79 N)^2) ≈ 777.92 N.
Direction of the resultant force, θ = arctan((Sum of y-components) / (Sum of x-components)).
θ = arctan(293.79 N / 717.71 N) ≈ 21.7°.
Therefore, the resultant force of the two forces is approximately 777.92 N at an angle of 21.7° with the forward dashed line.
(b) To find the acceleration of the car, we can use Newton's second law of motion, which states that the acceleration is equal to the net force divided by the mass of the object.
a = R / m.
Substituting the values we found for R and the given mass (m = 2950 kg):
a = (777.92 N) / (2950 kg) ≈ 0.264 m/s^2.
Therefore, the car has an acceleration of approximately 0.264 m/s^2.
To find the resultant of two forces, we can use vector addition. The resultant is the vector sum of the two forces.
(a) To find the resultant of the forces F1 and F2, we can break down the forces into their x and y components and then add the components separately.
The x-component of F1 can be calculated as F1x = F1 * cos(α), where α is the angle between the force and the x-axis.
F1x = 328 N * cos(35°) = 328 N * 0.8192 ≈ 268.83 N
The y-component of F1 can be calculated as F1y = F1 * sin(α), where α is the angle between the force and the x-axis.
F1y = 328 N * sin(35°) = 328 N * 0.5736 ≈ 188.49 N
Similarly, the x-component of F2 can be calculated as F2x = F2 * cos(β), where β is the angle between the force and the x-axis.
F2x = 464 N * cos(13°) = 464 N * 0.9744 ≈ 451.18 N
The y-component of F2 can be calculated as F2y = F2 * sin(β), where β is the angle between the force and the x-axis.
F2y = 464 N * sin(13°) = 464 N * 0.224951 ≈ 104.32 N
Now, we can add the x-components and y-components separately to find the resultant.
The x-component of the resultant force (Rx) can be calculated as Rx = F1x + F2x.
Rx = 268.83 N + 451.18 N ≈ 720.01 N
The y-component of the resultant force (Ry) can be calculated as Ry = F1y + F2y.
Ry = 188.49 N + 104.32 N ≈ 292.81 N
The magnitude of the resultant force (R) can be calculated as R = sqrt(Rx^2 + Ry^2).
R = sqrt((720.01 N)^2 + (292.81 N)^2) ≈ 781.07 N
The direction of the resultant force (θ) can be calculated as θ = tan^(-1)(Ry / Rx).
θ = tan^(-1)(292.81 N / 720.01 N) ≈ 22.57°
Therefore, the resultant of the two forces is approximately 781.07 N at an angle of approximately 22.57°.
(b) To calculate the acceleration of the car, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
F = m * a
Rearranging the formula, we can calculate the acceleration as:
a = F / m
Given that the mass of the car is 2950 kg and the magnitude of the resultant force (R) is approximately 781.07 N, we can find the acceleration of the car.
a = 781.07 N / 2950 kg ≈ 0.264 m/s^2
Therefore, the car has an acceleration of approximately 0.264 m/s^2.
a. Fr = 328N[35o] + 464N[13o]
X = 328*Cos35 + 464*Cos13 = 721 N.
Y = 328*sin35 + 464*sin13 = 293 N.
Fr^2 = X^2 + Y^2 = 721^2 + 293^2 = 605,690.
Fr = 778 N. = Resultant force.
b. a = F/m = 778/2950 = 0.264