# Tarzan wights 820N, swings from a cliff at the end of a 20m vine that hangs from a high tree limb and initially makes an angle of 22 degrees with the vertical. Immediately after Tarzan steps off the cliff the tension in the cine is 760N. Choose a coordinate system for which the x axis points horizontally away form the edge of the cliff and the y axis points up. a)What is the magnitude and direction of the net force acting on Tarzan? b) What is the magnitde and direction of his acceleration?

What are the equations?

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opps sorry but only the first part of my question was answered so I posted again so the other parts of the question would be answered as well.

You could have posted another note directly under the last post, as I am doing here. USUALLY we go back and look at our answers to see if there is a follow up questions or comment.

You could have posted another note directly under the last post, as I am doing here. USUALLY we go back and look at our answers to see if there is a follow up questions or comment.

## No worries! I apologize for any confusion. Let's continue with the remaining parts of your question.

To find the magnitude and direction of the net force acting on Tarzan, we need to consider both the gravitational force and the tension in the vine.

a) The gravitational force can be calculated using the formula F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity. In this case, the weight of Tarzan is given as 820N. Therefore, the force due to gravity is F_gravity = 820N.

Since the tension in the vine is not in the vertical direction, we need to resolve it into vertical and horizontal components. The vertical component of tension (T_y) counteracts the gravitational force, so T_y = F_gravity = 820N.

The horizontal component of tension (T_x) provides the net force acting on Tarzan in the x direction. To find T_x, we use the fact that the tension in the vine immediately after Tarzan steps off the cliff is 760N. We can use trigonometry to find the horizontal component:

T_x = Tension * cos(angle)
T_x = 760N * cos(22°)
T_x ≈ 711.19N

Therefore, the net force acting on Tarzan is the horizontal component, which is approximately 711.19N, in the direction away from the edge of the cliff.

b) To find the magnitude and direction of Tarzan's acceleration, we need to consider the net force and his mass. The net force can be found by combining the horizontal and vertical components of force:

Net Force = √(T_x^2 + T_y^2)
Net Force ≈ √(711.19N^2 + 820N^2)
Net Force ≈ 1054.15N

The direction of the net force can be determined using trigonometry:

θ = tan^(-1)(T_y / T_x)
θ = tan^(-1)(820N / 711.19N)
θ ≈ 49.49°

The magnitude of Tarzan's acceleration is then given by Newton's second law, F = ma:

a = Net Force / mass
a ≈ 1054.15N / m

Since the mass of Tarzan is not given in the question, we can't determine the numerical value of the acceleration. However, we know that the direction of the acceleration is the same as the direction of the net force, which is approximately 49.49° away from the edge of the cliff.

Please note that the above calculations assume no air resistance and neglect any other external forces acting on Tarzan. It also assumes Tarzan's mass remains constant throughout the motion.