# Physics\Trig Question

I'm having trouble figuring out how I'm supposed to find the force vectors for this problem. I know I'm supposed to use trig somehow but I can't wrap my head around this since I have no sides just angles.

• You could get the side you need by drawing a perpendicular to the ceiling, to the lug on your mass. Then you get two nice triangles, with a square angle. – An aedonist Sep 22 '17 at 15:26
• I see what you're saying. From there I would then use sin(55 degrees)*300N to get the side of the perpendicular imaginary line i just created. My question is since the 300-N is a downwards force would I make it (-1)(sin55)(300)? Edit: Actually no that would be wrong since 300N is not any side. I'm still confused – Samuel Uribe Sep 22 '17 at 15:30
• Your "sides" are the unknown forces, they must balance out the gravitational force of the mass. – Triatticus Sep 22 '17 at 15:35
• hint: draw vector $F_3$ representing the gravitational force. That's how you can find values for $F_1$ and $F_2$ – Vasya Sep 22 '17 at 15:37
• Vector F3 would be the vertical perpendicular to the axis going towards the 300N mass. I would assume that this vector F3 would be 300N as well right? – Samuel Uribe Sep 22 '17 at 15:39

Resolve the forces horizontally and vertically, which gives you a pair of simultaneous equations.

Vertically: $$F_1 \sin 30 + F_2 \sin 35 = 300$$

Horizontally: $$F_1 \cos 30 = F_2 \cos 35$$

Rearrange the second equation, to get $F_1$ in terms of $F_2$, and then substitute into the first equation

• This is a lot for me to wrap my head around. I understand the vertical part because it makes sense. The horizontal part though is confusing as to how they would equal – Samuel Uribe Sep 22 '17 at 15:42
• @SamuelUribe it's because the weight is not moving sideways at all, so the horizontal forces have to balance. – Neal Sep 22 '17 at 15:42
• I was able to get (F2(cos35))/cos30 +(F2(sin35))=300 is this what you meant? – Samuel Uribe Sep 22 '17 at 15:54
• Yes, but I think you are missing the sin30 – Henry L Sep 22 '17 at 15:58

Hint:

As there is static equilibrium, the sum of the three forces ($F_1$, $F_2$ and the weight) is zero.

Obviously, if you consider the projections of the three forces on some axis, the sum is still zero. The trick is to choose the axis in a clever way to get some simplification.

For instance, if you project on an axis that is perpendicular to $F_2$, the $F_2$ contribution vanishes and you get an equation like

$$F_1\cos\alpha=W\cos\beta$$ where $\alpha$ and $\beta$ are angles that you need to determine.

• This complicates things even more. I know there's some super easy way to solve this that doesn't require doing something like this that just requires a system of equations or something – Samuel Uribe Sep 22 '17 at 15:58
• No, it simplifies things as one of the unknown disappears and the answer is immediate. This is the super easy way. – Yves Daoust Sep 22 '17 at 16:36

I did what Henry L suggested and used my given knowledge of F1(sin30)+F2(sin35)=300 and F1(cos30)-F2(cos35)=0 to put one set of the system of equations in terms of just F2. The resulting equation was:

F2(cos(35)/cos(30))sin30+F2sin(35)=300 From that, I factored out F2 and got F2((cos(35)/cos(30))sin30+sin(35))=300 I moved everything inside of the parenthesis into the other side F2=300/((cos(35)/cos(30))sin30+sin(35)) The result was F2=286.67

From there I can apply the same logic to get F1. Once that is done I will have my sides and obtaining the vectors will be easy.

You can understand through pictures, Using geometry, Resolving vectors I know it's not perfect but In the above picture the vertical vector represents both the resolutions of $F_1$ and $F_2$

Now you could just equate the forces Therefore, $$F_1\cos(30)=F_2\cos(35)$$ $$F_1\sin(30)+F_2\sin(35)=300$$