Forces of a wrecking ball against a wall at rest.

[physicalConst]

Hello everyone, this is my first time posting to these forums, and my first time posting any online help regarding math. As you can imagine, this is a bit awkward and embarrassing for me. My textbook has no examples of the type of problem below and hours of googling has failed me. Hopefully someone would be kind enough to help me with my dilemma!

Question Description
A wrecking ball that has a mass of 200kg is resting against a wall. It hangs from a cable from the top of a crane that is touching the wall. The cable makes an angle of 9° with the wall. Ignore friction between the wall and ball.

Questions To Be Answered

1. Apply Newtons second law, but do not substitute the letters for numbers.
2. What is the magnitude and direction of the force of the wrecking ball on the wall?

As always I attempted to draw a free body diagram but became instantly stuck. From the picture I attached, you can see the two FBD I drew. In the first one I encountered a force which I could not label. I felt like it should be classified as weight, but that would give me two weight forces in differing directions.

http://db.tt/v3KiVQvS

 

[tiny-tim]

welcome to pf!

hello physicalConst! welcome to pf!

your top diagram (with four arrows), i don't understand at all

your bottom diagram is correct (with only 3 forces), except of course that weight is always vertical

 

[physicalConst]

Thanks for your response tiny-tim!

When I made the first diagram I was also confused, I couldn't see how it could be possibly right, but I though I missing something about the fundamentals of forces. So I drew up that second diagram, and although I knew that weight should be vertical, the whole pushing against the wall threw me. I think I over analyzed the whole problem.

 

[tiny-tim]

hi physicalConst!

(just got up :zzz:)

I think I over analyzed the whole problem.

yeah!

don't do that … mechanics really is usually quite simple!

weight is always vertical, reaction forces on a frictionless surface are always normal, and tension is always along the cable or string

(and if you can't think of a reason for a force … it isn't there!)​

 

[Nickie]

Hello and welcome to the forums! Don't worry, we are all here to learn and help each other out. Let's break down this problem step by step.

First, let's define the forces acting on the wrecking ball. We have the weight of the ball, which is a downward force due to gravity, and the tension force from the cable, which is pulling the ball towards the wall. The angle between the cable and the wall is given as 9°.

Now, let's apply Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In this case, the ball is at rest, so its acceleration is 0. Therefore, the net force on the ball must also be 0.

Next, let's resolve the forces into their components. The weight force can be resolved into a horizontal component and a vertical component. The horizontal component is equal to the weight force multiplied by the sine of the angle between the cable and the wall (9°). The vertical component is equal to the weight force multiplied by the cosine of the angle.

The tension force can also be resolved into horizontal and vertical components. The horizontal component is equal to the tension force multiplied by the cosine of the angle, and the vertical component is equal to the tension force multiplied by the sine of the angle.

Since the ball is at rest, the net force in the horizontal direction must be 0. This means that the horizontal components of the weight and tension forces must be equal and opposite. Therefore, we can set up the following equation:

T cos 9° = mg sin 9°

Where T is the tension force, m is the mass of the ball, and g is the acceleration due to gravity (9.8 m/s^2).

Solving for T, we get T = mg tan 9°. Plugging in the values, we get T = (200 kg)(9.8 m/s^2) tan 9° = 34.4 N.

Now, to find the magnitude and direction of the force of the wrecking ball on the wall, we can use the Pythagorean theorem. The magnitude of the force is equal to the square root of the sum of the squares of the horizontal and vertical components. We know the vertical component (34.4 N) and the horizontal component (mg cos 9°), so we can plug those values into the equation to get