Hi, I've been stuck on this for days! I'd love if I can get some help.

Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass 80kg ) is on a slope at 16∘ to the horizontal, but the lower climber (mass 65kg ) has gone over the edge to a steeper slope at 32∘.
a)Assuming frictionless ice and a massless rope, what is the acceleration of the pair?

b)The upper climber manages to stop the slide with an ice ax. After the climbers have come to a complete stop, what force must the ax exert against the ice?

What I've done:
a)Let T be tension
a: acceleration

First Climber
F=W+Tsin(16)
F=ma
W+Tsin(16)=ma
Tsin(16)=80a-(80*9.81)....eqn1

Second climber
F=W-Tsin(32)
F=ma
W-Tsin(32)=ma
rearranging:
Tsin(32)=(65*9.81)-65a....eqn2

eqn1/eqn2:(algebraic rearranging)
where a=9.81? (not sure if correct)
b)I'm not too sure how to go about this.

Answer:

a) The acceleration of the pair is 9.81 m/s2.

b) The force exerted by the ax against the ice is equal to the sum of the weights of the two climbers, 145 kg, multiplied by the acceleration of the pair, 9.81 m/s2. Therefore, the ax must exert a force of 1425.45 N against the ice.

To solve part a), we need to use the equations you derived for each climber and solve the system of equations simultaneously.

First, let's rewrite the equations using the given values:

Equation 1: T * sin(16°) = 80a - (80 * 9.81)
Equation 2: T * sin(32°) = (65 * 9.81) - 65a

To solve the system of equations, we can use one of several methods, such as substitution or elimination. Let's use substitution.

Rearrange Equation 1 to solve for T in terms of a:

T = (80a - (80 * 9.81)) / sin(16°)

We can substitute this expression for T in Equation 2:

(80a - (80 * 9.81)) / sin(16°) * sin(32°) = (65 * 9.81) - 65a

Now, let's simplify this equation:

(80a - (80 * 9.81)) / sin(16°) * sin(32°) = 65 * 9.81 - 65a

Cross-multiply and combine like terms:

(80a - (80 * 9.81)) * sin(32°) = (65 * 9.81 - 65a) * sin(16°)

Expand and simplify:

80a * sin(32°) - (80 * 9.81) * sin(32°) = 65 * 9.81 * sin(16°) - 65a * sin(16°)

Now, let's isolate the term with 'a' on one side:

80a * sin(32°) + 65a * sin(16°) = 65 * 9.81 * sin(16°) + (80 * 9.81) * sin(32°)

Factor out 'a':

a * (80 * sin(32°) + 65 * sin(16°)) = 65 * 9.81 * sin(16°) + (80 * 9.81) * sin(32°)

Divide both sides by (80 * sin(32°) + 65 * sin(16°)) to solve for 'a':

a = (65 * 9.81 * sin(16°) + (80 * 9.81) * sin(32°)) / (80 * sin(32°) + 65 * sin(16°))

Now, substitute the given values and calculate 'a':

a ≈ (65 * 9.81 * sin(16°) + (80 * 9.81) * sin(32°)) / (80 * sin(32°) + 65 * sin(16°))

a ≈ 6.91 m/s² (rounded to two decimal places)

So, the acceleration of the pair is approximately 6.91 m/s².

Now, let's move on to part b).

To find the force the upper climber's ax must exert against the ice to stop the slide, we can use Newton's second law:

Force net = mass * acceleration

The net force acting on the upper climber is the force exerted by the ax minus the force of gravity. So, we can write:

Force ax - Force gravity = mass * acceleration

Since the upper climber stops sliding, the acceleration is zero. Therefore:

Force ax - Force gravity = 0

Rearrange the equation to solve for Force ax:

Force ax = Force gravity

The force of gravity is given by:

Force gravity = mass * gravity

Substituting the given values:

Force gravity = 80 kg * 9.81 m/s²

Solve for Force gravity:

Force gravity ≈ 784.8 N

Therefore, the force the upper climber's ax must exert against the ice to stop the slide is approximately 784.8 Newtons.

To find the acceleration of the two climbers, you correctly set up equations using Newton's second law (F = ma) for each climber:

First Climber:
F = W + Tsin(16)
F = ma

Second Climber:
F = W - Tsin(32)
F = ma

Here, F represents the net force acting on each climber, W is the weight of each climber (m * g, where m is mass and g is acceleration due to gravity), T is the tension in the rope, and a is the acceleration of the system.

Now, let's solve these equations step by step:

For the first climber:
W + Tsin(16) = ma

Substituting values:
80 * 9.81 + Tsin(16) = 80 * a

For the second climber:
W - Tsin(32) = ma

Substituting values:
65 * 9.81 - Tsin(32) = 65 * a

Since the rope is assumed to be massless and there is no friction, the tension in the rope is the same for both climbers. We can express Tsin(16) in terms of Tsin(32) by using the fact that the sum of the forces in the vertical direction should be zero:

Tsin(16) + Tsin(32) = 0

Simplifying this equation:
sin(16) = - sin(32)
- sin(16) = sin(32)

By using this relationship, we can substitute Tsin(16) in terms of Tsin(32) in the first climber's equation:

80 * 9.81 - Tsin(32) = 80 * a

Now we have two equations and two unknowns (Tsin(32) and a). We need to solve these equations simultaneously to find the values.

To find the force exerted by the ice ax to stop the slide (part b), we need to consider the climber's motion when coming to a stop. At that point, the net force acting on the climbers is zero. The force exerted by the ax on the upper climber will be equal in magnitude and opposite in direction to the force the ice exerts on the ax.

So, to find the force exerted by the ax against the ice, you can use the equation:

Force_ax = Force_ice

Since Force_ice is the force required to bring the climbers to a stop, it can be found by multiplying the mass of the system (80kg + 65kg = 145kg) by the acceleration the climbers experience (which you already found in part a).

Plug in the value of the acceleration you found in part a to calculate the force exerted by the ax against the ice.