Three students converge on the last textbook in the store. All three grab it and begin to pull in the directions shown in the above figure. Student A exerts a force of 77.64 N, student B pulls with a force of 18.52 N, and student C pulls with a force of 66.43 N at an angle θ(theta)=70.49 degrees above the -x axis.
Considering these three forces (ignore the force of gravity on the book), what is the magnitude of the net force on the book?
To find the magnitude of the net force on the book, we need to add up the three forces acting on it.
The force exerted by Student A is 77.64 N.
The force exerted by Student B is 18.52 N.
The force exerted by Student C is 66.43 N at an angle of 70.49 degrees above the -x axis.
To add up these forces, we first need to break down the force exerted by Student C into its x and y components.
The x-component of the force is given by Fx = F * cos(θ), where F is the magnitude of the force and θ is the angle.
Fx = 66.43 N * cos(70.49 degrees) = 66.43 N * 0.3423 = 22.76 N
The y-component of the force is given by Fy = F * sin(θ).
Fy = 66.43 N * sin(70.49 degrees) = 66.43 N * 0.9397 = 62.47 N
Now we can add up the forces in the x-direction and the y-direction separately.
The net force in the x-direction is the sum of the x-components of the forces exerted by the students.
Net Force in the x-direction = Force A (77.64 N) + Force B (0 N, as it doesn't have an x-component) + Force C (22.76 N) = 77.64 N + 22.76 N = 100.4 N
The net force in the y-direction is the sum of the y-components of the forces exerted by the students.
Net Force in the y-direction = Force A (0 N, as it doesn't have a y-component) + Force B (18.52 N) + Force C (62.47 N) = 18.52 N + 62.47 N = 81.99 N
Finally, we can find the magnitude of the net force by using the Pythagorean theorem.
Magnitude of the net force = √(Net Force in the x-direction^2 + Net Force in the y-direction^2)
Magnitude of the net force = √(100.4^2 + 81.99^2) = √(10,080.16 + 6,732.60) = √16,812.76 = 129.72 N
Therefore, the magnitude of the net force on the book is approximately 129.72 Newtons.
To find the magnitude of the net force on the book, we can resolve the given forces into their x and y components. Let's start with student A's force.
The x-component of student A's force can be found using the formula:
F_Ax = F_A * cos(theta_A)
where F_A is the magnitude of student A's force and theta_A is the angle between the force and the -x axis.
Substituting the given values, we have:
F_A = 77.64 N
theta_A = 0 degrees (since it is along the -x axis)
F_Ax = 77.64 N * cos(0) = 77.64 N
Similarly, the y-component of student A's force is:
F_Ay = F_A * sin(theta_A)
F_Ay = 77.64 N * sin(0) = 0 N
Now let's find the x and y components of student B's force.
The x-component of student B's force can be found using the formula:
F_Bx = F_B * cos(theta_B)
where F_B is the magnitude of student B's force and theta_B is the angle between the force and the -x axis.
Substituting the given values, we have:
F_B = 18.52 N
theta_B = 90 degrees (since it is perpendicular to the -x axis)
F_Bx = 18.52 N * cos(90) = 0 N
Similarly, the y-component of student B's force is:
F_By = F_B * sin(theta_B)
F_By = 18.52 N * sin(90) = 18.52 N
Finally, let's find the x and y components of student C's force.
The x-component of student C's force can be found using the formula:
F_Cx = F_C * cos(theta_C)
where F_C is the magnitude of student C's force and theta_C is the angle between the force and the -x axis.
Substituting the given values, we have:
F_C = 66.43 N
theta_C = 70.49 degrees
F_Cx = 66.43 N * cos(70.49) = 33.765 N
Similarly, the y-component of student C's force is:
F_Cy = F_C * sin(theta_C)
F_Cy = 66.43 N * sin(70.49) = 61.317 N
Now, to find the net force on the book, we add all the x-components and y-components together.
The net x-component of force (ΣF_x) = F_Ax + F_Bx + F_Cx = 77.64 N + 0 N + 33.765 N = 111.405 N
The net y-component of force (ΣF_y) = F_Ay + F_By + F_Cy = 0 N + 18.52 N + 61.317 N = 79.837 N
The magnitude of the net force (ΣF) can then be calculated using the Pythagorean theorem:
ΣF = sqrt(ΣF_x^2 + ΣF_y^2)
= sqrt((111.405 N)^2 + (79.837 N)^2)
≈ sqrt(12422.940 N^2 + 6376.791 N^2)
≈ sqrt(18799.731 N^2)
≈ 137.136 N
Therefore, the magnitude of the net force on the book is approximately 137.136 N.
resolve the three forces into x- and y-components.
Add them up, and take the magnitude.