[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?

[raheem94]

(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?

[ztibor]

(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

[Julii92]

(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.