# Having lots of difficulty with this statics+equilibrium problem.

• EtherMD
In summary, to find the minimum coefficient of static friction between a 43.4 kg ladder and the ground, the force of friction and normal force must be isolated by setting the sum of torques to 0. For a 200.00 kg horizontal beam supported by a cable at 13.3 degrees to the vertical, the angle of the force at the hinge can be calculated by finding the sum of angles in a triangle. The forces acting on the ladder include downward force from a second mass, horizontal component of force of gravity on the second mass, force of gravity on the ladder, normal force on the second mass, normal force on the ladder, and frictional force from the ground. The forces acting on the beam include gravity
EtherMD

## Homework Statement

A 43.4 kg, 3.2 m uniform ladder leans against a frictionless wall.* A 88.6 kg person is standing on the ladder down 1.18 m from the top of the ladder.* The ladder makes an angle of 53 degrees with the horizontal.* What is the minimum coefficient of static friction between the ladder and the ground so that the ladder does not slip?

## Homework Equations

$$\sum$$Fx=0
$$\sum$$Fy=0
$$\sum$$$$\tau$$=0
Fgx = mgsin$$\vartheta$$
Fgy = mgcos$$\vartheta$$
Ff = $$\mu$$*(normal force)

## The Attempt at a Solution

So the way I saw this was, if you have the person on the ladder and it begins sliding down the vertical (toward the left on the horizontal) then the force of friction would inadvertently have to be to the right. (opposing the direction of motion). To find the coffecient of static friction, I'll need the value for the force of friction and the normal force. I am unclear on how to find the force of friction but I am assuming it can be isolated out when doing the $$\tau$$ = 0 step.

## Homework Statement

A 200.00 kg uniform, horizontal beam is hinged at one end and at the other is supported by a cable that is at 13.3 degrees to the vertical.* * The beam is 2.00 m long.* Calculate the direction of the force at the hinge (measured with respect to the horizontal.

## Homework Equations

Sum of all angles in a triangle = 180.

## The Attempt at a Solution

So the vertical is perpendicular to the horizontal beam, making an angle of 90 degrees. We also know that the angle between the vertical and the supporting cable is 13.3 degrees. As such, 180-90-13.3 = 76.7.

Is this correct reasoning?

Last edited:
EtherMD said:

So the vertical is perpendicular to the horizontal beam, making an angle of 90 degrees. We also know that the angle between the vertical and the supporting cable is 13.3 degrees. As such, 180-90-13.3 = 76.7.

Is this correct reasoning?

First : What are the forces that act upon the beam? Make a list. And a drawing(no need to post it).

Last edited:
Lok said:
What cable? There is none in the problem but must be a typo :P.

First : What are the forces that act upon the ladder? Make a list. And a drawing(no need to post it).

It was for the second question, but for the first. The forces acting upon the ladder are:

Since the mass resting on the ladder is at an incline:
-Downward force from the second mass (vertical component for force of gravity acting on second mass)
-Horizontal component for the force of gravity acting on the second mass
-Force of gravity acting on the ladder itself
-Normal force acting on second mass (?)
-Frictional force from ground onto the ladder

EtherMD said:
It was for the second question, but for the first. The forces acting upon the ladder are:

Since the mass resting on the ladder is at an incline:
-Downward force from the second mass (vertical component for force of gravity acting on second mass)
-Horizontal component for the force of gravity acting on the second mass
-Force of gravity acting on the ladder itself
-Normal force acting on second mass (?)
-Frictional force from ground onto the ladder

yeah must have been blind, tried to edit but you were faster :P

There are of course to Normal forces acting on the ladder. Wall and ground. But good job.

So for the beam there is as above :
-gravity
-Normal force ( hinge )
-Tension ( beam )
-Tension (cable) or call it a Force.

Your reasoning is correct as both ends suffer the same forces, but the final angle is it up or down?

## 1. Why is this statics and equilibrium problem difficult?

There could be several reasons why this particular problem is difficult for you. It could be due to a lack of understanding of the underlying concepts, not having enough practice, or not being familiar with the problem-solving techniques. It is important to identify the specific areas that are causing difficulty and address them accordingly.

## 2. How can I improve my understanding of statics and equilibrium?

One of the best ways to improve your understanding is to practice solving problems. Start with simple problems and gradually move on to more complex ones. It is also helpful to review the underlying concepts and equations and make sure you understand them fully. Additionally, seeking help from a tutor or professor can also be beneficial.

## 3. What are some common mistakes to avoid in statics and equilibrium problems?

Some common mistakes to avoid include not carefully labeling forces and directions, not considering all forces acting on the object, and not using the correct equations or applying them incorrectly. It is important to carefully analyze the problem and double-check your work to avoid these mistakes.

## 4. How can I approach a statics and equilibrium problem?

A good approach to solving these problems is to first draw a clear diagram, labeling all forces and their directions. Then, identify all known and unknown quantities and choose the appropriate equations to solve for the unknowns. It is also helpful to break down the problem into smaller, more manageable parts.

## 5. Are there any resources that can help me with statics and equilibrium problems?

Yes, there are many resources available to help you with these types of problems. Your textbook, class notes, and online tutorials can provide a good foundation. You can also find practice problems and solutions online or seek help from a tutor or professor. Additionally, joining a study group with classmates can be beneficial.

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