The resultant of the force(-4i+8j)N and the force F gives an object of mass 6kg an acceleration of (2i+3j)m/s^2. Calculate the angle between F and the unit vector j.

F=ma

Since 6a = 12i+18j
and F•a = |F|*|a| cosθ
you can now find cosθ, and hence θ

16i+10j

(-16i -8j)

73.9 To calculate the angle between the force vector F and the unit vector j, we first need to find the components of the force F.

Given that the resultant force is the sum of the given force and F, and it equals (2i + 3j) m/s^2, we can write the equation:

(-4i + 8j) + F = 2i + 3j

Simplifying this equation, we can equate the coefficients of the i and j terms:

-4 + F(i) = 2(i)
8 + F(j) = 3(j)

From these equations, we can find the values of F(i) and F(j):

F(i) = 2(i) + 4
F(j) = 3(j) - 8

To calculate the angle between F and the unit vector j, we can find the dot product between the force vector F and the unit vector j, and then use the formula:

cos(theta) = dot_product / (magnitude_F * magnitude_j)

The dot product between F and j is given by:

dot_product = F(i) * 0 + F(j) * 1 = F(j)

To find the magnitudes of F and j, we can use the Pythagorean theorem:

magnitude_F = sqrt(F(i)^2 + F(j)^2)
magnitude_j = sqrt(0^2 + 1^2) = 1

Substituting these values into the formula for cosine, we have:

cos(theta) = F(j) / (magnitude_F * magnitude_j)

To find the angle theta, we can take the inverse cosine (arccos) of cos(theta):

theta = arccos[F(j) / (magnitude_F * magnitude_j)]

Now, we can substitute the values we calculated earlier to find the angle between F and the unit vector j.