Sum of torques/friction problem

[ddn87]

Homework Statement

What is the magnitude of the horizontal
A uniform 7.1 m tall aluminum ladder is leaing against a frictionless vertical wall. The
ladder has a weight of 227 N. The ladder slips
when it makes a 52.0◦ angle with the horizontal floor.
Determine the coefficient of static friction
between the ladder and the floor

Homework Equations

sum of torquest ET=0 and Newtons second EF=ma

The Attempt at a Solution

We did a problem like this in class today but we didnt take the coefficient of friction into account I am not sure where to start with this... I have drawn a diagram... and since It slips at 52 degrees i believe that is kinetic friction... rather than the static friction coefficient which i am trying to find out. If i go thought the sum of torques equation i find out that the force force of friction in the x direction should equal the normal force as the ladder pushes against the wall... could some one please help me understand this concept a little better?

 

[Doc Al]

and since It slips at 52 degrees i believe that is kinetic friction... rather than the static friction coefficient which i am trying to find out.

No, all you need to worry about is static friction. The ladder is at the point where it is just about to slip, so static friction is at its maximum.

If i go thought the sum of torques equation i find out that the force force of friction in the x direction should equal the normal force as the ladder pushes against the wall... could some one please help me understand this concept a little better?

Apply all the conditions for equilibrium: ΣF = 0 and ΣT = 0.

Draw a free body diagram of the ladder.

 

[kerimek]

As a scientist, it is important to approach problems like this by breaking them down into smaller components and utilizing known equations and principles to find a solution. In this problem, we are given the height and weight of the ladder, as well as the angle at which it slips. We also know that the ladder is leaning against a frictionless wall.

To find the coefficient of static friction, we can use the equation EF=ma, where F is the force of friction, m is the mass of the ladder, and a is the acceleration due to gravity. We can also use the equation ET=0, where T is the torque and the sum of all torques must equal zero.

First, let's consider the forces acting on the ladder. We have the weight of the ladder acting downward, and the normal force from the floor acting upward. Since the ladder is not moving vertically, the sum of these forces must be equal: Fg = Fn.

Next, let's consider the torques. The only torque acting on the ladder is the weight of the ladder, which is acting at the center of mass. We can use the equation T = Fd, where T is the torque, F is the force, and d is the distance from the pivot point (in this case, the bottom of the ladder) to the force. Since the ladder is not slipping, the sum of all torques must be equal to zero: T = 0. This means that Fd = 0, and since d cannot be zero, F must also be zero. This tells us that there is no friction acting on the ladder.

Since there is no friction, we can conclude that the coefficient of static friction between the ladder and the floor is zero. This makes sense, as the ladder is leaning against a frictionless wall and is not slipping horizontally.

In summary, by breaking down the problem into smaller components and using known equations and principles, we can determine that the coefficient of static friction in this scenario is zero. This understanding can help us in future problems involving similar situations.