Recent content by 09jml90

I think it was maybe a little bit of both. I had never taken calculus or physics in high school. I was really surprised at first at how to analyze problems. A little before mid semester I'm pretty sure I had a D.

I have been undecided for about 2 years now, and even took a year off from school because I didn't know what I wanted to do. I still can't decide. One thing that I am quite interested in is Astronomy, and even if I do not try to go to grad school for astronomy, I still feel like a physics major...

Sorry, the computer says I finally got it right. I don't know why I kept getting caught up in thinking about the other objects instead of just block 1. You told me that I had calculated the net force when I multiplied block 1 (9.4kg)(1.95m/s^2) = 18.33 N, which is = ma. The weight of block 1 was...

Thank you for your help everybody.

Ok. Breaking the forces down into components and using the given angle. The x component of tension must equal -7.0 N, and using this I could solve the y component (21.54 N) and ultimately solve the hypotenuse which was the tension. The tension turned out to be 22.65 N. Because it is in...

If you are allowed to offer me any more hints, here is what I've come to until I'm stuck. In analyzing the top block, the force going upwards is 18.33 N (9.4kg)(1.95m/s^2). There is a force going downwards, the weight, which I'm not sure about. Is the weight needed for only block 1 or the...

Never-mind that is wrong too.

So, block 1 has a force going up which is 184.9 N. The force going down is weight. Is weight here just the weight of block 1 or the rest of the system of the rope and block 2. If so, the weight of the rope and block 2 has a force of 61.936 N. And the block going up has a force of 184.9 N. So...

What kind of martial arts do you train in/interested in?

Do I assume the mass of the rope to be included with the mass of the first block? to give the force going up as (9.4kg + 1.42kg)(1.95 m/s^2) = 21.1 N

I found the tension of the bottom of the rope to be 57.6 N, and I'm working on the top part.

Because the ball is steady with no force giving it an increase in acceleration should all the forces cancel each other out for equilibrium?

Would the second force be weight?, w=mg (w=(4.9kg)(9.8m/s^2) and then possibly T-48.02=(mass of M1 + Mrope)(11.76 m/s^2) ?

Homework Statement In the diagram below, the horizontal force, F, holds the ball steady. Given F = 7.0 N, and \theta = 18 deg, calculate the mass of the ball. I can't get the link to the picture to work, but it looks like a basic free body diagram with a ball. The ball is m, with an arrow, F...

Homework Statement Two blocks m1= 9.4 kg and m2 = 4.9 kg are connected by a homogeneous rope that has a mass of mr = 1.42 kg. A constant vertical force, F = 184.9 N, is applied to the upper block. What is the magnitude of the acceleration of the system? What is the magnitude of the...