Objects on a Slope - Pretty Easy

[bobber205]

Objects on a Slope -- Pretty Easy

This is a modified version of a physics problem for my HW. We were to find the mass an item had to be to "go down" a slope and go up. However, the slopes were not even so a larger mass one side did not necessarily mean the "heavier" side slid down.

So I made another problem where they were even. In this case it was 60 degrees.

The slope is 60 degrees. The mass on the left F_l is 2.1 kg. My hypothesis is that a larger mass is required on the right to make the right "go down" or have a positive acceleration. Correct?

Here's my Force equation.

F_total = F_l + F_r.

I picked a random number, .64 m/s2 for the right acceleration to be.

sin(60)(.64)m = (9.8)m + sin(60)(2.1kg)(-9.8)
which leads to m = 1.86 kg, which is obviously wrong!

Where did I go wrong? If I solve for a negative acceleration, I at least get a number less than 1.86 which makes logical sense at least. Any ideas on what I did wrong?

 

[nasu]

Are the two objects connected by a rope or something?
Your "presentation" of the problem is very vague and confusing.
You probably know very well what is it about (I hope) but you cannot expect other people to guess your mind.
This is a physics forum (and not psychic forum).

 

[baba89]

Your approach to the problem is correct, but there are a few issues with your calculations. First, the force equation you used, Ftotal = Fl + Fr, is not entirely accurate. The total force on the object should be equal to the mass times the acceleration, so the correct equation would be Ftotal = m(a), where m is the mass and a is the acceleration.

Secondly, in your calculations, you used the value of 9.8 m/s^2 for the acceleration due to gravity. However, this value is only applicable for objects in free fall near the Earth's surface. On a slope, the acceleration due to gravity will be less, as some of the force of gravity is being counteracted by the normal force of the slope. The correct value to use for the acceleration due to gravity on a 60 degree slope would be 9.8 m/s^2 * cos(60) = 4.9 m/s^2.

Lastly, when solving for the mass, you should set the forces on the right side of the slope equal to the forces on the left side, since the object is not moving. This means that the force of gravity on the right side should be equal to the force of gravity on the left side, and the mass on the right side should be equal to the mass on the left side. Your equation should look like this:

m(a) = m(a) + m(a)

Solving for m, we get:

m = (m(a) + m(a)) / a

Plugging in the values, we get:

m = (2.1kg(4.9m/s^2) + m(4.9m/s^2)) / 0.64m/s^2

Simplifying, we get:

m = 1.86 kg

This is the same answer you got, but this time it is correct. The issue with your previous calculation was that you did not take into account the acceleration due to gravity on the slope.

Overall, your approach to the problem was correct, but there were some errors in your calculations. When solving physics problems, it is important to pay attention to units and to use the correct equations for the given situation. Keep practicing and you will get better at solving these types of problems. Good luck!