You are Here: Home >< Maths

# M1 resolving forces Watch

1. I hate resolving forces so much and when I thought I sussed it something comes up so can someone please help me out.

Two forces P and Q act on a particle at a point O. The force P has magnitude 15 N and the force Q has magnitude X newtons. The angle between P and Q is 150°, as shown in
Figure 1. The resultant of P and Q is R.

Given that the angle between R and Q is 50°, find
(a) the magnitude of R

I know that I have to resolve vertically but for resolving forces I was under the impression that the forces that point up equal those that point down. Like kind of the forces in different directions equal each other.

So I thought it would be 15sin30 + Rsin30 = 0

However it's actually 15sin30 = Rsin30

ohhh confusion, please some help! I attached a good forces diagram to illustrate the information.

thanks
Attached Images

2. It's easiest to do this by a triangle of forces. Draw the force P of "length" 15 horizontal, then the force Q of length X at 150 degrees to the horizontal, so the angle between P and Q is 30. Then you can draw the force R from O to the end of Q; the angle between R and Q being 50 degrees. Now one can use the sine rule to find R.
3. I am generally fine with resolving it the standard way but I just don't get this concept of when to add or minus the forces.

I think for this one I am getting confused between whether the particle is at equilibrium or not. I just realised it doesn't say equilibrium. So I am I right to think if it's in equilibrium all the forces towards teh right = forces towards the left. However in the case of no equilibrium would it be X(unknown) - every other parallel force?
4. is the answer 9.79? if so i can help.
Ok just checked, i used the sin rule for my solution. To answer your question i dont think it matters which sign you use because the mag will always be positive. (i think you meant Rsin50 not R sin30*)
5. (Original post by T.P.D-L)
is the answer 9.79? if so i can help
Yeah it is.

I know I have to resolve vertically and I can get all of the horizontal forces however I just don't understand whether we add the two horizontal forces together ot minus them or make them equal to each other. I know the mark scheme makes them equal to each other but why?
6. (Original post by T.P.D-L)
is the answer 9.79? if so i can help.
Ok just checked, i used the sin rule for my solution. To answer your question i dont think it matters which sign you use because the mag will always be positive. (i think you meant Rsin50 not R sin30*)

Updated: May 21, 2009
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Grade boundaries not out until results day

What does this mean for you?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams