What happens to the force on the rope A-C as the weight changes? O The magnitude stays the same and the angle stays the same. O The magnitude decreases and theta changes. O The magnitude increases and the angle stays the same. The magnitude stays the same and the angle changes.

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What happens to the force on the rope A-C as the weight changes?
The magnitude stays the same and the angle stays the same.
The magnitude decreases and theta changes.
The magnitude increases and the angle stays the same.
The magnitude stays the same and the angle changes.
Transcribed Image Text:What happens to the force on the rope A-C as the weight changes? The magnitude stays the same and the angle stays the same. The magnitude decreases and theta changes. The magnitude increases and the angle stays the same. The magnitude stays the same and the angle changes.
Coplanar forces occur within two dimensions (A plane).
In this activity, you will observe the following when the mass of the crate is changed:
1. The change in the smallest angle theta (8) without either cable failing.
2. The forces on the ropes A-C and A-B.
0
10
C
20
0
30
40
A B
---8
50
+ Fx = 0; -Fc cose + FB = 0;
0 = 0.00⁰
60
70
Fc=
10(10³N) =-
80 90
+1ΣFy = 0;
Fc sine -ON=0
Equation (2)
The maximum force in either rope AB or AC cannot exceed 10 KN without breaking. As seen in the
x-equilibrium equation, Fc is always > FB. Thus, setting Fc to 10 KN represents the load limit of the
ropes.
Substituting Fc = 10 kN into Equation (2), we get
[10 (10³)N] sine -0.00 N = 0
= sin ¹ (0.000) = 0.00⁰
The force developed in rope AB can be obtained by substituting the values for 8 and Fc into Equation (1).
FB
cos
FB
cos 0.00⁰
100 110
FB = 10.00 KN
Drag the slider to change the mass of the crate.
FD
- Equation (1)
120
=
130
F.
B
140 150
0.00 N
X
160
170 180 190 200
Transcribed Image Text:Coplanar forces occur within two dimensions (A plane). In this activity, you will observe the following when the mass of the crate is changed: 1. The change in the smallest angle theta (8) without either cable failing. 2. The forces on the ropes A-C and A-B. 0 10 C 20 0 30 40 A B ---8 50 + Fx = 0; -Fc cose + FB = 0; 0 = 0.00⁰ 60 70 Fc= 10(10³N) =- 80 90 +1ΣFy = 0; Fc sine -ON=0 Equation (2) The maximum force in either rope AB or AC cannot exceed 10 KN without breaking. As seen in the x-equilibrium equation, Fc is always > FB. Thus, setting Fc to 10 KN represents the load limit of the ropes. Substituting Fc = 10 kN into Equation (2), we get [10 (10³)N] sine -0.00 N = 0 = sin ¹ (0.000) = 0.00⁰ The force developed in rope AB can be obtained by substituting the values for 8 and Fc into Equation (1). FB cos FB cos 0.00⁰ 100 110 FB = 10.00 KN Drag the slider to change the mass of the crate. FD - Equation (1) 120 = 130 F. B 140 150 0.00 N X 160 170 180 190 200
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