Two forces are applied to a car in an effort to accelerate it, as shown below. The first force, F1 = 364 N, is applied at an angle α = 33° to the forward dashed line. The second force, F2 = 522 N, is applied at an angle β = 11° to the forward dashed line.
What is the resultant of these two forces?
in N and at what degree
If the car has a mass of 3050 kg, what acceleration does it have? (Disregard friction.)
in m/s2 at what degree
Well, isn't this a car-azy situation! Let's calculate the resultant force first.
To find the resultant force, we need to break down both forces into their x and y components. Don't worry, it's not as complicated as parallel parking!
First, let's deal with Force F1. The x component (Fx1) can be found using the formula Fx1 = F1 * cos(α), where α is the angle.
Fx1 = 364 N * cos(33°)
Fx1 = 304.06 N (approximately)
Similarly, the y component (Fy1) of Force F1 is given by Fy1 = F1 * sin(α), where sin stands for the sine function.
Fy1 = 364 N * sin(33°)
Fy1 = 194.4 N (approximately)
Now, let's tackle Force F2. Just like before, we'll find its x and y components.
Fx2 = 522 N * cos(11°)
Fx2 = 510.69 N (approximately)
Fy2 = 522 N * sin(11°)
Fy2 = 96.4 N (approximately)
To find the resultant force (Fres), we add up the x and y components separately.
Fresx = Fx1 + Fx2
Fresx = 304.06 N + 510.69 N
Fresx = 814.75 N (approximately)
Fresy = Fy1 + Fy2
Fresy = 194.4 N + 96.4 N
Fresy = 290.8 N (approximately)
Finally, with the x and y components of the resultant force in hand, we can find the magnitude (Fres) using the Pythagorean theorem:
Fres = √(Fresx^2 + Fresy^2)
Fres = √(814.75 N^2 + 290.8 N^2)
Fres = √((667,215.56 N^2 + 84,528 N^2))
Fres = √(751,743.56 N^2)
Fres = 867.9 N (approximately)
Oh, the suspense is killing me! The resultant force is approximately 867.9 N.
Now, let's shift gears and rev up the acceleration! We'll use Newton's second law, F = ma, to find the acceleration of the car. Since we know the resultant force (Fres) and the mass of the car (m = 3050 kg), we can rearrange the formula to solve for acceleration (a).
a = F / m
a = 867.9 N / 3050 kg
a = 0.2844 m/s² (approximately)
Hooray! The car accelerates at approximately 0.2844 m/s².
Now, for some extra fun, if you'd like to know at what degree the resultant force is acting, we can use trigonometry. We'll find the angle (θ) between the resultant force and the forward dashed line using the formula:
θ = tan⁻¹(Fresy / Fresx)
θ = tan⁻¹(290.8 N / 814.75 N)
θ = tan⁻¹(0.3569)
θ ≈ 20.3° (approximately)
The resultant force acts at an angle of approximately 20.3° with respect to the forward dashed line.
Zoom, zoom! Hope that answers your question and left a smile on your face!
To find the resultant of two forces, we can use vector addition.
1. Resolve each force into its horizontal and vertical components:
The horizontal component of F1, F1x, is given by F1x = F1 * cos(α)
F1x = 364 N * cos(33°) = 303.84 N
The vertical component of F1, F1y, is given by F1y = F1 * sin(α)
F1y = 364 N * sin(33°) = 191.47 N
Similarly, for F2:
F2x = 522 N * cos(11°) = 515.92 N
F2y = 522 N * sin(11°) = 92.34 N
2. Find the resultant horizontal and vertical components by adding the corresponding components:
Resultant horizontal component, Rx = F1x + F2x = 303.84 N + 515.92 N = 819.76 N
Resultant vertical component, Ry = F1y + F2y = 191.47 N + 92.34 N = 283.81 N
3. Use the Pythagorean theorem to find the magnitude of the resultant force:
Resultant force, R = √(Rx^2 + Ry^2)
R = √(819.76 N^2 + 283.81 N^2)
R = √(672,912.5 N^2)
R = 820.9 N (rounded to one decimal place)
4. Calculate the angle of the resultant force with respect to the horizontal:
Angle of the resultant force, θ = tan^(-1)(Ry/Rx)
θ = tan^(-1)(283.81 N/819.76 N)
θ ≈ 19.76° (rounded to two decimal places)
Therefore, the resultant of these two forces is approximately 820.9 N at an angle of 19.76°.
Now, let's move on to the second part of the question:
5. Calculate the acceleration of the car using Newton's second law, F = ma:
The total force acting on the car is equal to the resultant force:
820.9 N = (3050 kg) * a
Solving for acceleration:
a = 820.9 N / 3050 kg
a ≈ 0.269 m/s^2 (rounded to three decimal places)
6. Since the acceleration is in the same direction as the resultant force, the angle of the acceleration is the same as the angle of the resultant force: 19.76°.
Therefore, the car has an acceleration of approximately 0.269 m/s^2 at an angle of 19.76°.
To find the resultant of the two forces, we need to break each force into its x-component and y-component. Then we can add the x-components together and the y-components together to get the resultant vector.
First, let's find the x- and y-components of each force:
For F1:
Fx1 = F1 * cos(α)
= 364 N * cos(33°)
= 303.18 N
Fy1 = F1 * sin(α)
= 364 N * sin(33°)
= 193.94 N
For F2:
Fx2 = F2 * cos(β)
= 522 N * cos(11°)
= 515.07 N
Fy2 = F2 * sin(β)
= 522 N * sin(11°)
= 94.51 N
Now, we can add the x-components and y-components together to find the resultant:
Rx = Fx1 + Fx2
= 303.18 N + 515.07 N
= 818.25 N
Ry = Fy1 + Fy2
= 193.94 N + 94.51 N
= 288.45 N
To find the magnitude of the resultant vector, we can use the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
= sqrt((818.25 N)^2 + (288.45 N)^2)
= sqrt(669230.06 N^2)
= 818.94 N (rounded to two decimal places)
To find the angle of the resultant vector, we can use the inverse tangent:
θ = atan(Ry / Rx)
= atan(288.45 N / 818.25 N)
= atan(0.3526)
= 19.62° (rounded to two decimal places)
Therefore, the resultant of the two forces is approximately 818.94 N at an angle of 19.62°.
To find the acceleration of the car, we can use Newton's second law of motion:
F = m * a
R = m * a
a = R / m
where R is the magnitude of the resultant force and m is the mass of the car.
a = 818.94 N / 3050 kg
= 0.2685 m/s^2 (rounded to four decimal places)
Therefore, the car has an acceleration of approximately 0.2685 m/s^2.
Note: The degree mentioned in the question may refer to the direction of the acceleration, which is the same as the angle of the resultant force. In this case, the angle is 19.62°.
Take components of the force:
F1x = 364*cos(33)
F1y = 364*sin(33)
F2x = 522*cos(11)
F2y = 522*sin(11)
Fnetx = F1x + F2x
Fnety = F1y + F2y
Fnet = ((Fnetx)^2 + (Fnety)^2)^0.5
at an angle theta given by
tan(theta) = Fnety/Fnetx
where Fnet = m*a = 3050*a
Solve for a, the acceleration