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A Level maths (mechanics) help please? watch

1. Hi, could someone help with this vectors question please?

A particle P of mass 6kg moves under the action of two forces, F1 and F2, where F1 = (8i-10j) N and F2 = (pi+qi) N, p and q are constants. The acceleration of P is a = (3i-2j) ms^-2.

a) Find the angle between the acceleration and i.

b) Find the values of p and q.

Thank you
2. (Original post by e123c)
Hi, could someone help with this vectors question please?

A particle P of mass 6kg moves under the action of two forces, F1 and F2, where F1 = (8i-10j) N and F2 = (pi+qi) N, p and q are constants. The acceleration of P is a = (3i-2j) ms^-2.

a) Find the angle between the acceleration and i.

b) Find the values of p and q.

Thank you
Draw the vector of the acceleration on the i,j basis coordinate system. Then find the angle between it and i.

For part (b), use the fact that
3. Could be wrong but here's what i'd do:

a)i is basically the x-axis, so find the angle between vector a = (3i-2j) and the x-axis.

b) Resultant F = ma, therefore resultant of F1 and F2 = 6 x (3i-2j)
4. (Original post by RDKGames)
Draw the vector of the acceleration on the i,j basis coordinate system. Then find the angle between it and i.

For part (b), use the fact that
Brilliant, thanks!
5. (Original post by Phatty_Magoo)
Could be wrong but here's what i'd do:

a)i is basically the x-axis, so find the angle between vector a = (3i-2j) and the x-axis.

b) Resultant F = ma, therefore resultant of F1 and F2 = 6 x (3i-2j)
That's great, thanks!
6. For part A we are just looking at the triangle that is made by the acceleration components, it's best to draw the triangle out in a diagram but I will just describe it:
Acceleration is (3i-2j)ms^-2 so that measn if we plot i along the horizontal and j along the verticle, we have a triang that has a horizontal side 3 units to the right and a vertical side 2 units downwards, using Tan(x)=o/a we can find the angle where a=3 and o=2 so x=Tan^-1(2/3)=33.690067...=33.7 degrees but remember the triangle this creates actually goes down below i so the angle would actually be -33.7 degrees (certain exam boards want you to specify but a diagram showing where that angle is should be enough explanation)
For part B consider F=MA for this, first find the resultant force in terms of p and q as this is what causes the acceleration: F(resultant)=F1+F2=8i+pi-10j+qj=(8+p)i+(q-10)j
Now we can find the acceleration in terms of p and q: A=F/M so A=((8+p)i+(q-10)j)/6=(8+p)i/6+(q-10)i/6
We know that this equals (3i-2j)ms^-2 so we can create two equations from the i and j components: from the i component we get (8+p)/6=3 which gives p=10 when you solve it and from the j component we get (q-10)/6=-2 which gives q=-2 when you solve it, these are your answers.
7. (Original post by MattSull)
For part A we are just looking at the triangle that is made by the acceleration components, it's best to draw the triangle out in a diagram but I will just describe it:
Acceleration is (3i-2j)ms^-2 so that measn if we plot i along the horizontal and j along the verticle, we have a triang that has a horizontal side 3 units to the right and a vertical side 2 units downwards, using Tan(x)=o/a we can find the angle where a=3 and o=2 so x=Tan^-1(2/3)=33.690067...=33.7 degrees but remember the triangle this creates actually goes down below i so the angle would actually be -33.7 degrees (certain exam boards want you to specify but a diagram showing where that angle is should be enough explanation)
For part B consider F=MA for this, first find the resultant force in terms of p and q as this is what causes the acceleration: F(resultant)=F1+F2=8i+pi-10j+qj=(8+p)i+(q-10)j
Now we can find the acceleration in terms of p and q: A=F/M so A=((8+p)i+(q-10)j)/6=(8+p)i/6+(q-10)i/6
We know that this equals (3i-2j)ms^-2 so we can create two equations from the i and j components: from the i component we get (8+p)/6=3 which gives p=10 when you solve it and from the j component we get (q-10)/6=-2 which gives q=-2 when you solve it, these are your answers.
Thank you!

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