You are Here: Home

# Combining forces Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 20-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Combining forces
A particle of mass m moves in a straight line on a horizontal surface under the influence of horizontal forces (2mx, -m)*, (m, mx), (mx, 2mx). The particle has an acceleration of 2ms-2. Find the two possibe values of x.

*This is in vector notation

Using F=ma, the magnitude of the resultant force on the particle is 2m. I then tried to find the magnitudes of the component forces mentioned above, and equate them to 2m to find the value of x. I got:

4m2x2 + m2 + m2 +m2x2 + m2x2 + 4m2x2 = 4m2 ( I removed all the square root signs and squared the 2m to make the calculation easier)

However I end up with 5x2=1, with x=1/root5. Apparantly the answer is +-(1/3). Can anyone tell me where I went wrong?
2. Re: Combining forces
(Original post by Julii92)
A particle of mass m moves in a straight line on a horizontal surface under the influence of horizontal forces (2mx, -m)*, (m, mx), (mx, 2mx). The particle has an acceleration of 2ms-2. Find the two possibe values of x.

*This is in vector notation

Using F=ma, the magnitude of the resultant force on the particle is 2m. I then tried to find the magnitudes of the component forces mentioned above, and equate them to 2m to find the value of x. I got:

4m2x2 + m2 + m2 +m2x2 + m2x2 + 4m2x2 = 4m2 ( I removed all the square root signs and squared the 2m to make the calculation easier)

However I end up with 5x2=1, with x=1/root5. Apparantly the answer is +-(1/3). Can anyone tell me where I went wrong?
Is there a diagram given?

I don't understand the meaning of (2mx, -m)*, (m, mx), (mx, 2mx), can you type it in the way it is written in the question, using LaTex?
3. Re: Combining forces
(Original post by Julii92)
A particle of mass m moves in a straight line on a horizontal surface under the influence of horizontal forces (2mx, -m)*, (m, mx), (mx, 2mx). The particle has an acceleration of 2ms-2. Find the two possibe values of x.

*This is in vector notation
In vector notation these can not be horizontal forces when m>0
I assume the resultant force is.

Using F=ma, the magnitude of the resultant force on the particle is 2m. I then tried to find the magnitudes of the component forces mentioned above, and equate them to 2m to find the value of x. I got:
THe resultant force in your vector notation: (m+3mx, -m+3mx)
this will be horizontal, when:
a) -m+3mx=0 for x>=0
b) -m -3mx=0 for x<0
4. Re: Combining forces
(Original post by ztibor)
In vector notation these can not be horizontal forces when m>0
I assume the resultant force is.

THe resultant force in your vector notation: (m+3mx, -m+3mx)
this will be horizontal, when:
a) -m+3mx=0 for x>=0
b) -m -3mx=0 for x<0
Thanks, got the right answer now.