(Physics) Tension in a rope

Question

A 25kg box is pulled by a rope at angle of 25 degrees above the horizontal and slides across the floor with an acceleration of 1.5 m/s^2. 30N of force is also acting on the box vertically. What is the tension in the rope when the coefficient of kinetic friction between the floor and box is 0.35?

Answer 1

OK so I'd start with a free-body diagram

1. The block's weight, W or F_g is acting down. F_g = mg = 25*g
2. The 30 N force is also acting down
3. A normal force is acting upward, but we don't know how big it is.
4. Friction is acting backwards (I'll call it left I guess). We can't solve for that yet without F_normal, but we know for now that F_f = μ_k*F_normal = .35*F_normal
5. Tension is the tricky one we will break down into its components
6. You can do this by drawing the Tension vector at 25-degrees, making a right triangle, and finding the vertical and horizontal components
7. T_x = Tcos25
8. T_y = Tsin25

Now we can start trying to solve by looking at the x and y forces separately

1. y forces
2. These 4 forces will add up to 0, but the problem is that one of them involves T, which we don't know yet, and another is F_normal, which we also don't know.
3. 0 = Tsin25 + F_normal - 30 - 25g
4. x forces
5. There are 2 , but we don't know either again. They do not add up to 0, since the block is accelerating at 1.5 m/s², so the forces don't add up to 0
6. F_net = ma
7. Tcos25 - .35 F_normal = 25(1.5)

We are left with 2 equations with 2 unknowns. Since we are just trying to find T here, the easiest way is to get an expression for F_normal by itself, which will be in terms of T. Then we can plug that into the other equation and solve for T.

1. Using the first equation:
2. 0 = Tsin25 + F_normal - 30 - 25g
3. 30 + 25g - Tsin25 = F_normal
4. Plug this into the other
5. Tcos25 - .35 F_normal = 25(1.5)
6. Tcos25 - .35 (30+25g - Tsin25) = 25*1.5
7. I will leave the calculation to you.