Recent content by Remle

@kuruman @PeroK Yeah, sorry for not showing work. Here it is: If ##\text{vector-15}## is along the ##x\text{-axis}##, then: $$\theta = \sin^{-1} {(\frac {15 \sin{140}} {47})}$$ $$\theta = \text{28°}$$ If ##\text{vector-35}## is along the ##x\text{-axis}##, then: $$\theta = \sin^{-1} {(\frac {35...

Using your diagram and ##x\text{-axis}## for the reference angle I get 28 degrees. Am I right?

I know is asking only for the magnitude of the vector. I believe the question is, which vector connects head-to-tail to whom to get the angle from the ##x\text{-axis}##?

So, for this problem is difficult to say? Does it need to specify which force is the first one? I, most of the time, choose the ##x\text{-axis}## as a reference but that too gave me 12 degrees and 28 degrees.

Ok. My problem is what angle to choose when adding vector. Statement does not tell me which one is the "first" force vector. So, when using the law of sine formula I get two results. First, using cosine to get the magnitude: $$\vec c = \sqrt{a^2 + b^2 +2ab\cos\theta},$$ $$\vec c = \sqrt{15^2 +...

Sorry for the late response. For the distance since is a scalar just had to add all the numbers. ##d = 1.3 + 2.5 = 3.8~\rm{km}##.

THIS is what I needed. So first displacement is ##\Delta d = 1.3 - 0## and the second is ##\Delta d = 0 - 2.5## so to speak.

I having a little bit of problem with ##\Delta d = d_f - d_i##. When substituting fo ##d_f## and ##d_i##, should I follow the signs rule (positive or negative)? For example, The problem shows that the displacement is ##1.2~\rm{km}, south## by solving ##\Delta d = -2.5 + 1.3## and I get that, but...

OMG... feel dumb. The ## 200~N ## is already the diagonal force. So the only thing I have to do is find ## F_\rm{ay} ## and it add to ## F_\rm{g} ##. Then only force left is ## F_\rm{ax} ## which that gives me ## 100~N ## dividing that with ## 25~\rm{kg} ## provides the ## 4~\rm{m/s^2} ##...

$$F_\rm{a_\rm{y}} = -200 \times \sin(60) = -173.205~N$$ $$F_\rm{a_\rm{x}} = 200 \times \cos(60) = 100~N$$ $$F_\rm{g}= 25 \times -9.81=-245.25~N$$ $$F_\rm{Net_\rm{y}} = - 245.25 - 173.205 + 245.25 = -173.205~N$$ $$F_\rm{Net_\rm{x}} = 100~N$$ $$F_\rm{Net}=\sqrt{{(-173.205)^2} + 100^2} = 200.00~N$$...

The problem does not say anything about a surface, so I just assumed it because that's how a normal force works. When I say original is I calculate FN with mass and gravity. Then I start adding or substracting other forces (if there's any) to this FN. Fg=245.25N and Fay=-173.205 N

What is the acceleration of the box? Paper says the answer is 4 m/s2. What is the Normal force acting on the box? Paper says the answer is 418 N. I know that for most cases FN=Fg=W. So, by definition the "original" Normal force is 245.25 N (am I correct?) I calculated the Fay which is...

First of all, thanks for accepting me! Math teacher by profession, but recently I've been getting into the area of physics. PF has helped me in many situations so I've decided to finally join the club and keep learning.