Two identical boxes of masses 45. Both experiences a sliding force with mew of 0.15. Find the tension in the chord and the acceleration of the boxes. The angle of inclination is 30°

Answer 1

To solve this problem, we need to consider the forces acting on the boxes. The forces include gravitational force pulling the boxes downwards (mg), the normal force acting perpendicular to the incline (N), the tension in the chord (T), and the force of friction (f).

Since the boxes are identical, the tension in the chord will be the same for both boxes. Let's denote this tension as T.

Let's break down the forces on each box:
- The gravitational force pulling the boxes downwards can be broken down into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ).
- The normal force acting perpendicular to the incline will be equal to the perpendicular component of the gravitational force, which is mg*cosθ.
- The force of friction is given by f = μ*N, where N is the normal force and μ is the coefficient of friction. In this case, f = 0.15*(mg*cosθ).
- The net force acting on the boxes parallel to the incline is T - mg*sinθ - f.
- The net force is also equal to the mass times acceleration.

Since the boxes are connected by a chord, the tension in the chord will accelerate both boxes. Therefore, the net force acting on both boxes is equal to the total mass of both boxes times acceleration.

Setting up the equations for both boxes:

Box 1:
T - 45*g*sin(30°) - 0.15*(45*g*cos(30°)) = 45*a

Box 2:
T - 45*g*sin(30°) - 0.15*(45*g*cos(30°)) = 45*a

Since T is the same for both boxes, we can add the two equations together and solve for T:

2T - 90*g*sin(30°) - 0.15*(90*g*cos(30°)) = 90*a

Now, we can solve for T:

T = 90*g*sin(30°) + 0.15*(90*g*cos(30°))

Calculating T using g = 9.81 m/s^2, we get:

T = 508.126 N

Now we can solve for the acceleration (a) by substituting T back into one of the initial equations:

T - 45*g*sin(30°) - 0.15*(45*g*cos(30°)) = 45*a

508.126 - 45*9.81*sin(30°) - 0.15*(45*9.81*cos(30°)) = 45*a

a = 0.92 m/s^2

Therefore, the tension in the chord is 508.126 N and the acceleration of the boxes is 0.92 m/s^2.