You are Here: Home >< Maths

# M1 resolving forces

Announcements Posted on
Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher! 27-10-2016
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*)

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

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

### University open days

Is it worth going? Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams