1. The problem statement, all variables and given/known data I've been having a little bit of trouble answering this question: "An 8.0 kg wooden sled is pulled over the snow by means of a rope that makes an angle of 40° with the horizontal. If the rope has a tension of 70.0 N, what is the normal force acting on the sled?" 2. Relevant equations w = mg , gives the weight of an object FN = mg cos(θ) , gives the normal force acting on an object on an inclined surface. But since it hasn't been mentioned in my textbook yet, I'm thinking that the question has to be answered using vector components only. X = R cos(θ) , and: Y = R sin(θ) , gives the horizontal and vertical components of a vector respectively. 1. The problem statement, all variables and given/known data Unfortunately, I don't have a picture of my diagram, but what I first drew is a triangle with a 40° angle corner, representing the hill. I drew a vector pointing straight down, starting from the centre of the sled on the hill, marked "78.48 N", which is the weight of the sled. I drew another vector pointing upwards, perpendicular to the hill surface, starting from the centre of the sled, representing the normal force which needs to be solved. On the side of the sled, I drew a vector, parallel to the hypotenuse of the hill, marked "70 N", which is the tension on the rope. I formed a triangle using that side and, by checking, I found that the corner angle of this triangle was 50°. Lastly, above and beneath the diagram of the hill, I joined the weight vector, which acted as the hypotenuse, with the normal force vector, which acted as the adjacent side, forming a triangle. 3. The attempt at a solution First, I converted the mass of the object into its weight: w = mg w = 8.0 * - 9.81 w = -78.48 N This is where the problem began. Now that I had a vector triangle with the hypotenuse calculated to be -78.48 N and a corner angle of 40°, I used the normal force formula to find the adjacent side (the normal force excluding the tension of the rope). And since the calculated normal force excluded the tension of the rope, I was going to find the difference in magnitude between the calculated normal force and the Y component of the 70 N (the tension) triangle to find the normal force including the tension of the rope. So I found the normal force (excluding the tension of the rope): FN = mg cos(θ) FN = 8.0 * -9.81 cos(40) FN = -78.48 cos(40) FN = -60.12 Since the tension vector was the X component of the triangle I drew, I found the Y component of the triangle using trigonometry: tan(θ) = Opposite/Adjacent tan(50) = Opposite/70 Opposite = tan(50) 70 Opposite = 83.42 N Now that I had the Y component of the triangle, I found the difference between that and the calculated normal force: Normal Force Including Tension Vector = Calculated Normal Force + Y Component Of Tension Vector Normal Force Including Tension Vector = -60.12 + 83.42 Normal Force Including Tension Vector = 23.3 N So I found that the normal force was 23.3 N, but when I checked my textbook the answer was 33.4 N Maybe I was overcomplicating things and horribly confused. I'm struggling with this question and it would be great if somebody could help me and tell me where I went wrong. I will post a picture of my diagram later. Also, this is my first post on PF and so I'm really excited! (And upset about the question). Thank you.