What would the greatest force on the vine be during the swing?

[shell4987]

Homework Statement

Tarzan, who weighs 619 N, swings from a cliff at the end of a convenient vine that is 14 m long (Fig. 8-37). From the top of the cliff to the bottom of the swing, he descends by 4.3 m. The vine will break if the force on it exceeds 1550 N. What would the greatest force on the vine be during the swing?
**image attached**

Homework Equations

Fynet=T-mgcos(theta)=(mv^2)/r
T=mgcos(theta)+(mv^2)/r
0+mgh=(1/2)mv^2+0 (initial and final energies)
v^2=2gh

The Attempt at a Solution

I attempted this problem and plugged everything into the tension equation, but I don't know how to get the angle for the cosine, I put in zero and my tension was 424.44 N and that's obviously not right. Can anyone help me out?

 

[Astronuc]

The greatest force on the vine will occur at maximum tangential speed (and therefore maximim kinetic energy). The tension is the resultant the component of Tarzan's weight in the direction parallel with the vine and centrifugal force due to the circular motion of Tanzan's mass.

So with mgy = 1/2 mv² one can find v, where y is the vertical distance traveled.

 

[m2003]

As a scientist, the greatest force on the vine during the swing can be calculated by considering the forces acting on the vine at different points of the swing. At the top of the swing, the only force acting on the vine is the tension force, which is equal to the weight of Tarzan (619 N). As Tarzan swings down, the tension force decreases due to the decrease in the radius of the swing. At the bottom of the swing, the tension force is at its minimum and is equal to the weight of Tarzan minus the centripetal force (1550 N). This is the point at which the vine will break.

To find the greatest force on the vine, we can use the conservation of energy principle. At the top of the swing, all of Tarzan's potential energy (mgh) is converted into kinetic energy (1/2 mv^2). At the bottom of the swing, all of the kinetic energy is converted back into potential energy. Therefore, we can set the equations for potential energy at the top and bottom of the swing equal to each other:

mgh = (1/2)mv^2

Solving for v, we get:

v = sqrt(2gh)

Substituting this value for v into the equation for tension, we get:

T = mgcos(theta) + (mv^2)/r

Plugging in the given values of mass (m), gravity (g), and radius (r), we can solve for the tension (T) at the bottom of the swing:

T = 619 cos(theta) + (619^2)/(14)

Since we know that the vine will break if the tension exceeds 1550 N, we can set this equation equal to 1550 and solve for the angle (theta):

1550 = 619 cos(theta) + (619^2)/(14)

Solving for theta, we get:

theta = cos^-1(931/619) = 24.7 degrees

Therefore, the greatest force on the vine during the swing would occur at the bottom of the swing, when the tension force is equal to 1550 N and the angle between the vine and the vertical is 24.7 degrees.