Solve Kinetic Friction Homework: Acceleration & Time Before Rock Goes Over Edge

[orange03]

Homework Statement

A 75 kg climber finds himself dangling over the edge of an ice cliff, as shown in the figure below. Fortunately, he's roped to a 980 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.5×10−2. What is his acceleration, and how much time does he have before the rock goes over the edge? Neglect the rope's mass. What is his acceleration? How much time does he have before the rock goes over the edge?

Homework Equations

F=ma

The Attempt at a Solution

equation of the climber:
Ft-mg=may

equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N. I plugged this back into the climber's equation and got -2.76 m/s^2 but this answer is wrong. Help please!

 

[songoku]

equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N.

The equation of the rock is wrong
The tension is not the same as friction because the rock moves with acceleration.
Use $$\sum F$$ = ma to get the equation for the rock

 

[tomcue]

I would first like to commend you on your attempt at solving this problem. However, there are a few mistakes in your equations and approach that may have led to the incorrect answer. Let's break down the problem and see where we can make corrections.

First, let's define the forces acting on the climber and the rock. For the climber, we have the tension force from the rope pulling up, the weight force (mg) pulling down, and the friction force (Ff) acting in the opposite direction of motion. For the rock, we have the tension force pulling up and the weight force pulling down.

Now, let's set up the equations of motion for the climber and the rock. For the climber, we have:

ΣF = Ft - mg - Ff = may

For the rock, we have:

ΣF = Ft - mg = mar

We can see that the tension force (Ft) is present in both equations. This is because the climber and the rock are connected by the rope, so the tension force is the same for both of them. Now, let's find the values of the other forces.

The weight force for both the climber and the rock is simply their respective masses (m) multiplied by the acceleration due to gravity (g = 9.8 m/s^2).

The friction force is given by the equation Ff = μkN, where μk is the coefficient of kinetic friction and N is the normal force. The normal force in this case is equal to the weight force (mg) for both the climber and the rock.

Plugging in the values, we get:

For the climber: Ft - (75 kg)(9.8 m/s^2) - (0.055)(75 kg)(9.8 m/s^2) = (75 kg)a

For the rock: Ft - (980 kg)(9.8 m/s^2) = (980 kg)a

We can solve these equations simultaneously to find the value of the tension force (Ft) and the acceleration (a).

Next, we need to find the time before the rock goes over the edge. We can use the equation d = vot + 0.5at^2, where d is the distance (51 m), vo is the initial velocity (which we can assume is 0 m/s since the rock is initially at rest), a