?????Two books of mass m1 = 6.9 kg and m2 = 3.8 kg are stacked on a table as shown in the figure below. Find the normal force acting between the table and the book on the bottom.

SO g*m1+g*m2=104.86
N = mgcos(θ)
N= (104.86)(9.81)Cos?
Where do i find cos?
IS THIS RIGHT?

Try:

N=(m1*a + m1*g) + (m2*a + m2*g)

Also, since the books aren't moving the acceleration is 0.

Yes, you are on the right track. To find the normal force between the table and the book on the bottom, you need to use the equation N = mgcos(θ).

In this case, the value of θ is not given in the question, so we need to find it. The figure shown in the question is not visible in the text-based format, so we cannot determine the angle directly from the figure.

However, we can assume that the books are stacked vertically, which means θ would be 0 degrees (or π/2 radians) since the force due to gravity acts directly downwards. In this case, cos(0) = 1.

So, if you assume θ = 0 degrees, the equation becomes N = (104.86)(9.81)(cos(0)).

Since cos(0) = 1, the equation can be simplified to N = (104.86)(9.81)(1) = 1029.6846 N (approximately).

Therefore, the normal force acting between the table and the book on the bottom is approximately 1029.6846 N.

Please note that this assumption is made based on the given information in the question. If there are any additional details or a different scenario depicted by the figure, please consider those factors in determining the value of θ.