When dealing with stress components, we have to apply a sine convention. We will be using the following sine convention throughout this course. Now consider shear stress component tau xy. We know from the subscripts that this stress acts on a surface that has a normal factor in the x direction and the resultant force from that shear stress is acting in the y direction. To apply the sine to this stress, we will first take the magnitude of that stress. We will then multiply it by the direction of the surface normal relative to our coordinate system. So if the surface normal is in the positive x direction, we would multiply it by positive 1. We then multiply that by the sine for the resultant force direction. So if it's in the positive y direction, this would be a positive 1 value. And remember, if we're dealing with normal stresses, the single subscript can be expanded to a double subscript in order to determine the sine. Let's look at this for an example. Here we see two stress states that are the same in direction but opposite in sine. If we look at the stress state on the left, our outward surface normal is in the positive z direction and our normal stress sigma z as a resultant in the positive z direction as well. So our stress becomes sigma z magnitude of sigma z multiplied by positive 1 multiplied by positive 1, which gives us a positive normal stress. Remember, the first positive number is a sign of the outward normal and the second positive number is the sign of the resultant force. On the right, we have a positive outward normal direction but our resultant is in the negative direction. So in this case, sigma z would be negative. If we apply that to the shear stress components on the left example, we see that they are all in the positive direction. While in the right, they are all in the negative direction.