A dog tugs forward with force of 28n on a taut leash at an angle of 30 above the horizontal what is the magnitude of the horizontal component of the tension in the leash?

To find the horizontal component of the tension in the leash, you can use the cosine of the angle. The horizontal component ( F_x ) is calculated as ( F_x = F \cdot \cos(\theta) ), where ( F ) is the force (28 N) and ( \theta ) is the angle (30 degrees). Thus, ( F_x = 28 , \text{N} \cdot \cos(30^\circ) \approx 28 , \text{N} \cdot 0.866 \approx 24.9 , \text{N} ). Therefore, the magnitude of the horizontal component of the tension in the leash is approximately 24.9 N.