Tarzan, who weighs 820 N, swings from a cliff at the end of a 20.0 m vine that hangs from a high tree limb and initially makes an angle of 22.0° with the vertical. Assume that an x axis extends horizontally away from the cliff edge and a y axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 760 N. Just then, what are (a) the force on him from the vine in unit-vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the x axis? What are the (e) magnitude and (f) angle of Tarzan’s acceleration just then?
We can first find the x and y components of the tension force in the vine.
(a) T_x = T*cos(22°) = 760*cos(22°) and T_y = T*sin(22°) = 760*sin(22°).
Calculating these values, we get T_x = 719.4 N and T_y = 287.69 N. Thus, the force on Tarzan from the vine in unit-vector notation is F_T = 719.4 i + 287.69 j (N).
(b) Now we need to find the net force on Tarzan. Since Tarzan weighs 820 N, there is a gravitational force F_g = -820 j (N) acting on him. The net force is the vector sum of the tension force and the gravitational force: F_net = F_T + F_g.
F_net = (719.4 i + 287.69 j) + (0 i - 820 j) = 719.4 i - 532.31 j (N)
(c) To find the magnitude of the net force, we use the Pythagorean theorem: |F_net| = sqrt((719.4)^2 + (-532.31)^2) = 901.17 N
(d) To find the angle of the net force relative to the positive direction of the x-axis, we use the tangent function: tan(theta) = (-532.31) / (719.4).
Solving for theta, we find theta = atan(-532.31 / 719.4) = -36.5°.
(e) Tarzan's acceleration can be found using Newton's second law: F_net = m*a, where m is Tarzan's mass and a is the acceleration vector.
First, we need to find Tarzan's mass. Weight = mass * g (gravity), so mass = Weight / g = 820 N / 9.81 m/s^2 = 83.6 kg.
Now we can find Tarzan's acceleration components in x and y: a_x = F_net_x / m = 719.4 N / 83.6 kg = 8.6 m/s^2, and a_y = F_net_y / m = -532.31 N / 83.6 kg = -6.36 m/s^2.
The magnitude of Tarzan's acceleration is |a| = sqrt((8.6 m/s^2)^2 + (-6.36 m/s^2)^2) = 10.68 m/s^2.
(f) The angle of Tarzan's acceleration relative to the positive direction of the x-axis can be found using the tangent function again: tan(theta) = -6.36 / 8.6.
Solving for theta, we find theta = atan(-6.36 / 8.6) = -36.5° (the same as the angle of the net force).
To find the force on Tarzan from the vine and the net force on him, we can break down the forces into their x and y components.
(a) Force on Tarzan from the vine:
The force on Tarzan from the vine can be found by resolving the tension in the vine into its x and y components. The x component of the force is equal to the tension in the vine multiplied by the cosine of the angle between the tension and the x-axis, and the y component is equal to the tension multiplied by the sine of the angle.
Fx = Tension * cos(angle)
Fx = 760 N * cos(22.0°)
Fx ≈ 697.33 N
Fy = Tension * sin(angle)
Fy = 760 N * sin(22.0°)
Fy ≈ 274.51 N
The force on Tarzan from the vine can be represented in unit-vector notation as (697.33 N, 274.51 N).
(b) Net force on Tarzan:
The net force is the vector sum of all the forces acting on Tarzan. Since the only force acting on him is the force from the vine, the net force will be equal to the force from the vine.
The net force on Tarzan can be represented in unit-vector notation as (697.33 N, 274.51 N).
(c) Magnitude of the net force:
The magnitude of the net force can be found using the Pythagorean theorem.
Magnitude = sqrt(Fx² + Fy²)
Magnitude = sqrt((697.33 N)² + (274.51 N)²)
Magnitude ≈ 758.77 N
The magnitude of the net force on Tarzan is approximately 758.77 N.
(d) Angle relative to the positive x-axis:
The angle of the net force relative to the positive x-axis can be found using the inverse tangent function.
Angle = atan(Fy / Fx)
Angle = atan(274.51 N / 697.33 N)
Angle ≈ 21.46°
The angle of the net force on Tarzan relative to the positive direction of the x-axis is approximately 21.46°.
(e) Magnitude of Tarzan's acceleration:
The acceleration of Tarzan can be found using Newton's second law, F = ma, where F is the net force and a is the acceleration. Since the mass of Tarzan is not given, we will assume it cancels out in this case.
Magnitude of acceleration = Magnitude of net force
Magnitude of acceleration ≈ 758.77 N
The magnitude of Tarzan's acceleration is approximately 758.77 N.
(f) Angle of Tarzan's acceleration:
Since the net force and acceleration have the same magnitude, their angles will also be the same.
The angle of Tarzan's acceleration is approximately 21.46°.