Learning Goal: To use the conditions of equilibrium to solve for unknown forces

Question

Learning Goal: To use the conditions of equilibrium to solve for unknown forces in a three-dimensional force system. For particle equilibrium, the sum of all forces must be zero: sigma F = 0 If the forces are resolved into their i, j, and k components, the condition for equilibrium becomes sigma F_x i + sigma F_y j + sigma F_z k = 0 which results in the following three scalar component equations that can be used to solve for unknowns: sigma F_x = 0 sigma F_y = 0 sigma F_z = 0 Figure 2 Correct When using vectors, pay attention to the location of the vector's head and tail. F_x is pointing toward the origin rather than away from the origin. If a vector is defined by two arbitrary points A(x_A, y_A, z_A) and B (x_B, y_B, Z_B), where point A is the head and point B is the tail, the vector's resulting components are r = (x_A - x_y) i + (y_A - y_B)j + (z_A - z_B) k As shown, a system of cables supports a bucket. (Figure 2). The dimensions in the figure are as follows x_1 = 4.35 ft, x_2 = 1.70 ft, y_1 = 2.00 ft, y_2 = 4.00 ft, z_1 = 3.15 ft, and z_2 = 3.20 ft. If the bucket and its contents have a combined weight of W = 17.0 lb, determine T_A, T_B, and T_C, the tensions in cable segments DA, DB, and DC, respectively. Express your answers numerically in pounds to three significant figures separated by commas. T_A, T_B, T_c = 10.9, 2.42, 13.5 lb

Explanation / Answer

we need to determine the unit vectors along the three cables

Cable DB : unit vector=-j

cable DA : Coordinates of A=(4.35,0,3.15); Coordinates of D=(1.7,2.0,0)

vector along DA = 2.65i-2j-3.15k

Unit vector along DA = (2.65i-2j-3.15k)/4.577 = 0.579i-0.437j-0.688k

Cable DC : Coordinates of C=(0,4,3.2)

vector along DC = -1.7i+2j+3.2k

Unit vector along DC = -0.411i+0.483j+0.773k

Let tension magnitudes in three cables be TDB , TDA, TDC

In vector forem, the tensions are

TDB(-j) ; TDA(0.579i-0.437j-0.688k);TDC ( -0.411i+0.483j+0.773k)

External weight = -17k

For equilibrium of the bucket,

TDB + TDA+TDC =17k

Or,

TDB(-j) +  TDA(0.579i-0.437j-0.688k) + TDC ( -0.411i+0.483j+0.773k) = 17k

Or,

0.579TDA - 0.411TDC =0

-TDB-0.437TDA+0.483TDC =0

-0.688TDA+0.773TDC = 17

Solving above system of equations,we get

TDB = 10.32lb

TDA=59.73lb

TDC=42.4 lb

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.