Two children pull a third child on a snow saucer sled exerting forces
F1 and F2 as shown from above in Figure 4.36. Find the acceleration
of the 49.00-kg sled and child system. Note that the direction of the
frictional force is unspecified; it will be in the opposite direction of the sum
of F1=10N and F2=8 N
F1 have an angle 45 degree with horizontal line and F2 has an angle 30 degree below the horizontal line at fourth quadrant
also there is another force directed to the left on x axises = 7.5 N
Well, it seems like the sled and child have quite the party going on! Let's see if we can calculate their acceleration and join in on the fun.
First, we need to break down the forces acting on the sled and child system. We have F1, F2, and the force directed to the left on the x-axis. Let's call this force Fl.
Now, F1 has an angle of 45 degrees with the horizontal line, and F2 has an angle of 30 degrees below the horizontal line. Since we're in the fourth quadrant, we can conclude that these forces are both acting in the positive x-axis direction.
Given that F1 = 10 N and F2 = 8 N, we can calculate the x-component of each force using trigonometry. The x-component of F1 would be F1cos(45°), and the x-component of F2 would be F2cos(30°).
Using a calculator, we find that the x-component of F1 equals approximately 7.07 N, and the x-component of F2 equals approximately 6.93 N.
Now, let's sum up all the x-components of the forces acting on the sled and child system. Since we have a force directed to the left on the x-axis, we need to subtract its magnitude from the sum of F1 and F2.
The sum of the x-components would be (7.07 N + 6.93 N) - 7.5 N = 6.50 N.
Great! Now, we have the net force acting on the sled and child system in the x-direction. To find the acceleration, we can use Newton's second law: Fnet = ma.
The mass of the sled and child system is given as 49.00 kg. Let's plug in the values:
Fnet = ma
6.50 N = 49.00 kg * a
Now, let's solve for the acceleration a:
a = 6.50 N / 49.00 kg
Calculating this, we find that the acceleration of the sled and child system is approximately 0.1327 m/s^2.
So, the sled and child system are accelerating at a rate of approximately 0.1327 meters per second squared. Time to slide on over and join the wintery fun!
To find the acceleration of the sled and child system, we need to calculate the net force acting on the system.
Let's break down the forces acting on the sled and child system:
1. F1: magnitude = 10 N, angle = 45 degrees above the horizontal line.
2. F2: magnitude = 8 N, angle = 30 degrees below the horizontal line.
3. Frictional force: unknown magnitude, opposite direction to F1 + F2.
4. Force directed to the left on the x-axis: magnitude = 7.5 N.
First, let's resolve F1 and F2 into their x and y components:
F1x = F1 * cos(45 degrees)
= 10 N * cos(45 degrees)
= 10 N * √2/2
≈ 7.07 N
F1y = F1 * sin(45 degrees)
= 10 N * sin(45 degrees)
= 10 N * √2/2
≈ 7.07 N
F2x = F2 * cos(-30 degrees)
= 8 N * cos(-30 degrees)
= 8 N * √3/2
≈ 6.93 N
F2y = F2 * sin(-30 degrees)
= 8 N * sin(-30 degrees)
= 8 N * (-1/2)
= -4 N
Now, let's calculate the net force in the x-direction:
Net Force in the x-direction = F1x + F2x + Force to the left
= 7.07 N + 6.93 N + (-7.5 N)
= 6.57 N
Now, let's calculate the net force in the y-direction:
Net Force in the y-direction = F1y + F2y
= 7.07 N + (-4 N)
= 3.07 N
Since the frictional force is in the opposite direction of the net force, and we know that the net force in the x-direction is to the left, the frictional force must be directed to the right.
Now, let's calculate the net force magnitude:
Net Force magnitude = sqrt((Net Force in the x-direction)^2 + (Net Force in the y-direction)^2)
= sqrt((6.57 N)^2 + (3.07 N)^2)
≈ sqrt(43.1049 N^2 + 9.4249 N^2)
≈ sqrt(52.5298 N^2)
≈ 7.253 N
The mass of the sled and child system is given as 49.00 kg.
Finally, we can find the acceleration using Newton's second law:
Net Force = mass * acceleration
Solving for acceleration:
acceleration = Net Force / mass
= 7.253 N / 49.00 kg
≈ 0.148 m/s^2
Therefore, the acceleration of the sled and child system is approximately 0.148 m/s^2.
To find the acceleration of the sled and child system, we need to consider the net force acting on the system.
First, let's resolve the given forces into their horizontal and vertical components:
F1 = 10 N at a 45-degree angle with the horizontal line
F2 = 8 N at a 30-degree angle below the horizontal line in the fourth quadrant
Force directed to the left on the x-axis = 7.5 N
Resolving F1:
F1_horizontal = F1 * cos(45°)
F1_horizontal = 10 N * cos(45°)
F1_horizontal ≈ 7.07 N (rounded to two decimal places)
F1_vertical = F1 * sin(45°)
F1_vertical = 10 N * sin(45°)
F1_vertical ≈ 7.07 N (rounded to two decimal places)
Resolving F2:
F2_horizontal = F2 * cos(-30°) (negative because it is below the horizontal line)
F2_horizontal = 8 N * cos(-30°)
F2_horizontal ≈ 6.93 N (rounded to two decimal places)
F2_vertical = F2 * sin(-30°) (negative because it is below the horizontal line)
F2_vertical = 8 N * sin(-30°)
F2_vertical ≈ -4 N (rounded to two decimal places)
Now, let's calculate the net forces in the horizontal and vertical direction:
Net horizontal force (F_horizontal_net):
F_horizontal_net = F1_horizontal + F2_horizontal + Force directed to the left on the x-axis
F_horizontal_net = 7.07 N + 6.93 N - 7.5 N
F_horizontal_net = 6.5 N
Net vertical force (F_vertical_net):
F_vertical_net = F1_vertical + F2_vertical
F_vertical_net = 7.07 N - 4 N
F_vertical_net = 3.07 N
The frictional force, which is in the opposite direction of the sum of F1 and F2, would be given as:
Frictional force = -(F1 + F2) = -(10 N + 8 N) = -18 N
Now that we have the net forces in the horizontal and vertical directions, we can calculate the acceleration using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
Net force = mass * acceleration
Using the net horizontal force:
F_horizontal_net = mass * acceleration_horizontal
6.5 N = 49.00 kg * acceleration_horizontal
So, the acceleration in the horizontal direction (acceleration_horizontal) is:
acceleration_horizontal = 6.5 N / 49.00 kg
acceleration_horizontal ≈ 0.133 N/kg (rounded to three decimal places)
Using the net vertical force:
F_vertical_net = mass * acceleration_vertical
3.07 N = 49.00 kg * acceleration_vertical
So, the acceleration in the vertical direction (acceleration_vertical) is:
acceleration_vertical = 3.07 N / 49.00 kg
acceleration_vertical ≈ 0.063 N/kg (rounded to three decimal places)
Finally, we can calculate the magnitude of the acceleration of the sled and child system using the Pythagorean theorem:
acceleration = √(acceleration_horizontal^2 + acceleration_vertical^2)
acceleration ≈ √((0.133 N/kg)^2 + (0.063 N/kg)^2)
acceleration ≈ √(0.0513 N^2/kg^2 + 0.00397 N^2/kg^2)
acceleration ≈ √0.0553 N^2/kg^2
acceleration ≈ 0.235 N/kg (rounded to three decimal places)
Therefore, the acceleration of the sled and child system is approximately 0.235 N/kg.
F = 10N[45o] + 8N[330o] -7.5N
X = 10*cos45 + 8*cos330 - 7.5 = 6.50 N.
Y = 10*sin45 + 8*sin330 = 3.07 N.
Tan A = Y/X = 3.07/6.50 = 0.47247
A = 25.29o
Fr = X/cosA = 6.50/cos25.29=7.19N[25.29]
= Resultant force.
a = Fr/m = 7.19[25.29o]/49=0.147[25.29]