w2526123
Alright thank you again. I have another question if you don't mind. I just couldn't ask my teacher, uhm it has something to do with normal forces. The formula is F=ma right? is acceleration due to gravity only used if there's no specific acceleration given regarding to a specific problem?
kharmsorz
Let’s say for example: the force F₁ has the angle of 30 degrees at the first quadrant whereas the other force F₂ has the angle of 55 degrees at the second quadrant. Thus, the x-components are: +F₁cos30° and –F₂cos55° The y-components are: +F₁sin30° and +F₂sin55°
kharmsorz
Consider our example that the angles of the forces F₁ and F₂ are both reference angles.
w2526123
Fair enough. And in able to calculate them, there must be a tension or forces applied right?
kharmsorz
The symbol “a” from the formula F=ma is the acceleration of the body due to the applied force. It is the same concept with the weight which is w=mg but this one acts only a vertical (downward) motion. If the situation is in equilibrium, meaning the body is not moving no matter how many applied forces, then you apply the Newton’s FIRST law of motion, and so the acceleration “a” is zero, but the weight w=mg still exists.
kharmsorz
Yes, there are tension or applied forces acting on the body. You will realize what the Newton’s FIRST and SECOND laws of motion all about. To make it our explanation short, if the NET FORCE is ZERO, then the body is in equilibrium (so acceleration is zero); if the NET FORCE is NOT ZERO, then the body moves (so there is an acceleration).
kharmsorz
If you need something more to be cleared up, don’t hesitate to ask. We can slowly chát here. But for now I’ll do elsewhere. Wish you all the best!
w2526123
Now I have the clearer explanation about w=mg is. It's like from a problem that our class solved was like Fn= -Fg then the next part was Fn= -mg ( that example problem has something to do with coefficient of friction now). I just knew that "weight" in physics is the force of gravity and "mass" is how heavy or light an object. I apologize for taking so long to respond. I simply had to run some errands. Thank you so much for you help, I greatly appreciate it!
Answers & Comments
See the attached image for the solution.
Thus, the x-components are:
+F₁cos30° and –F₂cos55°
The y-components are:
+F₁sin30° and +F₂sin55°