![omg please help i really dont get thisa) Show that the downward force F exerted by the clamp on the loaded cage is about 6*10^4N [4 marks]b) Calculate the initi class=](https://us-static.z-dn.net/files/d2b/cb024aea8cd29b83027373f866ea0ba6.jpg)
Answer :
Explanation:
a) To find the downward reaction force on the clamp, we simply sum the forces in the y direction.
∑F = ma
2T cos θ − F = 0
F = 2T cos θ
F = 2 (3.7×10⁴ N) (cos 20°)
F ≈ 7.0×10⁴ N
b) Sum the forces in the y direction again, only this time, there's no downward reaction force, and the acceleration isn't 0.
∑F = ma
2T cos θ = ma
a = (2T cos θ) / m
a = (7.0×10⁴ N) / (1.2×10³ kg)
a ≈ 58 m/s²
The sum of the forces acting on a body in equilibrium is zero
(a) The downward force exerted by the clamp, F = [tex]\mathbf{F_{Rope}}[/tex] - W is approximately 6.0 × 10⁴ N
(b) The initial acceleration of the loaded cage is approximately 48.128 m/s²
The process of arriving at the above values is as follows:
Known parameters;
The number of elastic ropes = 2
Tension in each elastic rope before release, T = 3.7 × 10⁴ N
Angle each rope makes with the vertical tower, θ = 20°
Total mass of the loaded cage, M = 1.2 × 10³ kg
The required parameter;
(a) To show that the downward force F exerted by the clamp is about 6 × 10⁴ N
Method;
Calculate the resultant force exerted by the clamp
Solution;
The upward forces due to the elastic rope, [tex]\mathbf{F_{Rope}}[/tex] = 2 × T × cos(θ)
∴ [tex]F_{Rope}[/tex] = 2 × 3.7 × 10⁴ × cos(20°) ≈ 69,537.25 N
The downward force due to the weight of the loaded mass, W = M × g
∴ W = 1.2 × 10³ kg × 9.82 m/s² = 11,784 N
The downward force F exerted by the clamp on the loaded cage, F, is given as follows;
F = [tex]\mathbf{F_{Rope}}[/tex] - W
∴ F = 69,537.25 N - 11,784 N = 57,753.25 N = 5.775325 × 10⁴ N ≈ 6.0 × 10⁴ N
The downward force F exerted by the clamp, F ≈ 6.0 × 10⁴ N
b) Required; The initial acceleration of the loaded cage
Method;
Newton's third law of motion, Force, F = Mass, M × Acceleration, a, can be applied
Solution;
Acceleration, a = F/M
The force acting on the loaded cage when the clamp is released is given by the downward force exerted by the clamp, F = 5.775325 × 10⁴ N
The acceleration of the loaded cage of mass, 1.2 × 10³ kg, is therefore;
[tex]a = \mathbf{ \dfrac{5.775325 \times 10^4}{1.2 \times 10^3 }} \approx 48.128[/tex]
The initial acceleration of the loaded cage when the clamp is released, a ≈ 48.128 m/s²
Learn more about equilibrium of forces here;
https://brainly.com/question/12582625