M1 Question

Hi guys... I'm wondering if someone could help me out with this question....

A force R acts on a particle, where R = (7i + 16j)N
Calculate

a) the magnitude of R to 1 d.p

b) the angle between the line of action of R and i, giving answer to nearest degree.

c) The force R is the resultant of the two forces P and Q. The line of action of P is parallel to the vector (i+4j) and the line of action of Q is parallel to the vector (i+j)

Determine the forces P and Q expressing each in terms of i and j.

a) I got 17.5 N for this which was right.
b) I got 66 degrees which was right.
c) the answers are p = (3i+12j) and Q = (4i+4j)... I just dont understand how to get these...

I thought that because they were parallel |p| = k(i+4j) and |q| = k(i+j)

I don't get integer values for k...unless they've rounded the answers..... could someone help out please?....

Reply 1

oceane

If P is parallel to the vector (i + 4j) it means it's a multiple of this vector. ie it's pointing in this direction but it could be any length so:

P = k (i + 4j) where k is a constant.
Similarly for Q;
Q = c(i + j) where c is a constant.

Now, since R is the resultant of both of these:
R = k(i+4j) + c(i+j)

You know that R is 7i + 16j, so grouping the i's and the j's:
i's 7=k + c
j's 16=4k + c which are simultaneous equations.

16-7 = 4k-k
9 = 3k and therefore k=3 and c = 4 when you substitute k into the simultaneous eqtn.

So P = 3i +12j
Q=4i + 4j

Reply 2

super_baros

Oh i see... I was trying to work with the magnitudes.... :confused:

..
and also I thought that the constant was the same for both of them...

Thanks!... you rock!...

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