Distance between tourques/forces

[natedigity]

In Figure 12-42 a 57 kg rock climber is in a lie-back climb along a fissure, with hands pulling on one side of the fissure and feet pressed against the opposite side. The fissure has width w = 0.15 m, and the center of mass of the climber is a horizontal distance d = 0.50 m from the fissure. The coefficient of static friction between hands and rock is μ1 = 0.45, and between boots and rock it is μ2 = 1.25. The climber adjusts the vertical distance h between hands and feet until the (identical) pull by the hands and push by the feet is the least that keeps him from slipping down the fissure. (He is on the verge of sliding.)
What is the value of h?

mg of the climber is 558.6N

What is the magnitude of the climber’s pull by his hands and push by his shoes? This is the same question as one asking for the magnitude of the normal force at his hands and shoes when he is on the verge of sliding.

Number 328.588 Units N


I have found the tourques using the foot/rock contact as my pivot point.

What are the torques due to (a) the frictional force on the hands, (b) the frictional force on the shoes, (c) the normal force on the shoes, and (d) the gravitational force on the climber, taken to act at the com? Include a minus sign if a torque is negative.

(a) Number -22.1797 Units N•m
(b) Number 0 Units N•m
(c) Number 0 Units N•m
(d) Number 363.09 Units N•m


The torque due to the climber’s pull by the hands on the fissure wall? That is, what is the torque due to the normal force on his hands? This is -340.91 Nm

I am confused on how to find the distance h as in the height in the y direction with the above torques. I am sure I am missing something stupid. Thanks, Nathan.

 

[Doc Al]

By just taking torques about that pivot point you are left with two unknowns: the normal force and h. You'll need another equation. What other conditions for equilibrium apply here?

 

[ogb p]

I would like to clarify that the distance h in this scenario refers to the vertical distance between the climber's hands and feet. This distance is important because it determines the magnitude of the normal force at his hands and feet, which in turn affects the frictional forces and torques involved in the climb.

To find the value of h, we need to first understand that the climber is on the verge of sliding, which means that the frictional forces at his hands and feet are equal to the maximum static friction that can be exerted by the rock. This can be represented by the equation μN = F, where μ is the coefficient of static friction, N is the normal force, and F is the maximum frictional force.

We can use this equation to find the normal force at the climber's feet, which is equal to the force exerted by his feet on the rock. This can be represented by the equation N = mg, where m is the mass of the climber and g is the acceleration due to gravity. Substituting the values given in the problem, we get N = (57 kg)(9.8 m/s^2) = 558.6 N.

Now, using the equation μN = F, we can find the maximum frictional force at the climber's hands, which is equal to the force exerted by his hands on the rock. This can be represented by the equation F = μN = (0.45)(558.6 N) = 251.37 N.

Since the climber is on the verge of sliding, the magnitude of the pull by his hands and the push by his feet must be equal to the maximum frictional force, which is 251.37 N. This means that the magnitude of the normal force at his hands must also be equal to 251.37 N.

Now, we can use this information to find the value of h. We know that the vertical distance between the climber's hands and feet is equal to the sum of the distances between his center of mass and his hands (d) and between his center of mass and his feet (h). This can be represented by the equation h + d = 0.50 m. Rearranging this equation, we get h = 0.50 m - d.

Substituting the value of d = 0.50 m and solving for h, we get h = 0