a. Define the term equilibrium b. The system shown in Figure Q1b below is in equilibrium. Calculate the…

Question

a. Define the term equilibrium b. The system shown in Figure Q1b below is in equilibrium. Calculate the values of the tensions in the strings A, B, C and D.

ScanSolve AI Tutor · Accepted Answer

Here's the solution to the problem.

a) Equilibrium is a state where the net force and net torque acting on an object are both zero. This means the object is either at rest (static equilibrium) or moving with constant velocity (dynamic equilibrium).

b) To calculate the tensions in the strings, we will analyze the forces at point C. The system is in equilibrium, so the sum of forces in both the horizontal and vertical directions must be zero.

Let $T_C$ be the tension in the string segment from the left wall to point C, and $T_D$ be the tension in the string segment from the right wall to point C. The angles are given as $20^\circ$ with the horizontal for $T_C$ and $35^\circ$ with the horizontal for $T_D$ . The downward force is $400 \, \text{N}$ .

Resolve forces horizontally.
The sum of horizontal forces is zero.
$\sum F_x = 0$ $T_D \cos(35^\circ) - T_C \cos(20^\circ) = 0$ $T_D \cos(35^\circ) = T_C \cos(20^\circ)$ $T_D = T_C \frac{\cos(20^\circ)}{\cos(35^\circ)}$ $T_D = T_C \frac{0.9397}{0.8192}$ $T_D = 1.1471 T_C \quad (*)$
Resolve forces vertically.
The sum of vertical forces is zero.
$\sum F_y = 0$ $T_D \sin(35^\circ) + T_C \sin(20^\circ) - 400 \, \text{N} = 0$ $T_D \sin(35^\circ) + T_C \sin(20^\circ) = 400 \, \text{N}$
Substitute $T_D$ from Step 1 into the vertical force equation.
$(1.1471 T_C) \sin(35^\circ) + T_C \sin(20^\circ) = 400 \, \text{N}$ $1.1471 T_C (0.5736) + T_C (0.3420) = 400 \, \text{N}$ $0.6580 T_C + 0.3420 T_C = 400 \, \text{N}$ $1.0000 T_C = 400 \, \text{N}$ $T_C = 400 \, \text{N}$
Calculate $T_D$ using the value of $T_C$ .
Substitute $T_C = 400 \, \text{N}$ into equation $(*)$ :
$T_D = 1.1471 \times 400 \, \text{N}$ $T_D = 458.84 \, \text{N}$
Determine tensions in strings A and B.
String A is the vertical segment supporting the horizontal string B. String B is the horizontal segment connected to point C.
The tension in string B is equal to the horizontal component of $T_C$ .
$T_B = T_C \cos(20^\circ)$ $T_B = 400 \, \text{N} \times \cos(20^\circ)$ $T_B = 400 \, \text{N} \times 0.9397$ $T_B = 375.88 \, \text{N}$
The tension in string A is equal to the tension in string B, as it is supporting the horizontal string B.
$T_A = T_B$ $T_A = 375.88 \, \text{N}$
The tensions are:
$\boxed{T_A = 375.88 \, \text{N}}$ $\boxed{T_B = 375.88 \, \text{N}}$ $\boxed{T_C = 400.00 \, \text{N}}$ $\boxed{T_D = 458.84 \, \text{N}}$

A. Define the term equilibrium b. The system shown in Figure Q1b below is in equilibrium. Calculate the values of the tensions in the strings A, B, C and D.

Step-by-step solution

Resolve forces horizontally.

Resolve forces vertically.

Substitute $T_D$ from Step 1 into the vertical force equation.

Calculate $T_D$ using the value of $T_C$ .

Determine tensions in strings A and B.

Still stuck on this one?

More Physics Questions