Two identical boxes of masses 45kg both experience a sliding force of U=0.15. Find the tension in the chord and the acceleration of the boxes. Angle is 30°
To find the tension in the chord and the acceleration of the boxes, we can use Newton's second law of motion:
1. Sum of forces in the x-direction:
T - 0.15 = 45a1....(1)
T - 45(9.8)sinθ = 45a2....(2)
where T is the tension in the chord, a1 is the acceleration of the first box, a2 is the acceleration of the second box, and θ = 30°.
2. Sum of forces in the y-direction:
45(9.8)cos30° = 45g
3. Solve equation 2 for T:
T = 45(9.8)sinθ + 45a2
4. Substitute T from equation 3 into equation 1:
45(9.8)sinθ + 45a2 - 0.15 = 45a1
5. Substitute sin(30°) = 0.5 into equation 4:
45(9.8)(0.5) + 45a2 - 0.15 = 45a1
222.75 + 45a2 - 0.15 = 45a1
6. Substitute cos(30°) = √3/2 into equation 2, equation 1, and sum them:
45(9.8)(√3/2) = 45g = T + 45(9.8)sin(30)
T - 0.15 = 45a1
T - 45(9.8)(√3/2) = 45a2
T - 45(9.8)(√3/2) = 45(9.8)(0.5) + 45a2 - 0.15
T = 110.25 + 45a2 - 0.15
7. Equate T from equations 3 and 6:
45(9.8)sinθ + 45a2 = 110.25 + 45a2 - 0.15
45(9.8)(0.5) = 110.25 - 0.15
222.75 = 110.1
222.75 - 110.1 = 112.65 = 45a2
a2 = 2.37 m/s^2
Therefore, the tension in the chord, T = 222.75 - 0.15 = 222.6 N
and the acceleration of the boxes, a2 = 2.37 m/s^2.