Recent content by Nova_Chr0n0

The questions and relevant formulas/information are attached below: I started by writing the needed values: 2 kN = 2,000 N 6378 km =6378(1000) m 1800 km = 1800(1000)m After converting, I tried solving for the mass using the law of gravitation formula: I decided to replace N as...

Sorry, I am really slow to catch up, so I need more clarification. My current understanding, based on the explanation, is that: The normal force produced by the roller is considered an internal force and not an external force. Is it because the roller is part of the whole figure and therefore...

The problem is from Hibeller's book, Mechanics: Statics and attached below is the picture of the problem: My question about this problem is about the FBD of the reactions. Here is how I drew it: But when I tried checking the solution for the problem, they have this as their FBD: My...

The figure and formulas is shown above. My strategy of cutting the areas/shapes is shown below: Area 1 = Area of Triangle Area 2 = Area of the square - Area of the quarter circle Area 3 = Area of the larger quarter circle - Area of the smaller quarter circle Computing for the areas, I got...

The solution that is at the last page of the e-book I've got (Mechanics - Hibeller 14th edition) also used moment at B. Here is its solution: But I've already got the correct solution, as Sir Chestermiller stated that I've left out one force.

Thank you very much! I didn't notice that I left out the component force of 250 N that is parallel to the bar. I've now gotten the same answer.

The figure is shown below: Here is my FBD for the figure with assign +x and +y directions I started off by summing up the forces in the x-direction: Next is the summing up of the forces in the y-direction: After this, I solved for the moment at point A: assuming that counter-clockwise is +...

Thanks! Upon reviewing it, my diagram is actually wrong. I have now corrected it and the resultant drawing using parallelgoram method now lands on the upper part of the x-axis. I'm also not a fan of using parallelogram espsecially when resultants are involved. But well, I got to follow the...

Picture above is the complete question. I want to ask about the problem where I would use the parallelogram method. Here is my FBD: I start off by computing the angle alpha: α + α + 105 +105 = 360 α = 75 degree After that, I now use cosine law to solve for the resultant force...

Now I understand it much better. Thanks for the explanation!

I've already got the answer and the way to solve it (parallelogram), but I'm just wondering why I cannot use the technique I've learned in the lesson torque. Let's focus on the line AB, if I use what I've learned in torque, the components would be like this: To find the force component in...

Is it the whole surface of the cylindrical vessel and not only its base area?

I've already got the correct answer in letter (a), which is 17140.2 Pascals. My question will be focusing about the letter b of the question and here is my solution: (b) FORMULA: P = F/A F = P*A My understanding about this problem is I have to use the pressure that I got in letter (a) to...

I guess there was just an error in the given answer for the file that I've encountered. Also, thanks for letting me know a solution with less work when involving triangle ratios. Really appreciate it!

There was no additional information about the problem; I just found a worksheet pdf file to practice torque problems. I guess there was just an error in the answer. If you're curious about it, here is the link...