Chapter 12: Q15P (page 346)

Question: Forces ${\vec{F}}_{1}, {\vec{F}}_{2} and {\vec{F}}_{3}$ act on the structure of Fig. 12-33, shown in an overhead view. We wish to put the structure in equilibrium by applying a fourth force, at a point such as P. The fourth force has vector components ${\vec{F}}_{h} and {\vec{F}}_{v}$ . We are given that a = 2.0 m,b = 3.0m , c = 1 0 m , ${\vec{F}}_{1} = 20 N, {\vec{F}}_{2} = 10 Nand {\vec{F}}_{3} = 5.0 N$ Find (a) F_h , (b) F_v, and (c) d.

Short Answer

Expert verified

Answer

$a . F_{h} = 5.0 N b . F_{v} = 30 N c . d = 1.3 m$

Step by step solution

Understanding the given information

$a = 2.0 m b = 3.0 m c = 1.0 m F_{1} = 20 N F_{2} = 10 N F_{3} = 5.0 N$

Concept and formula used in the given question

By applying the equations of static equilibrium, you can get the equations in terms of unknown forces. By solving these equations, you can find the values of unknown forces and distance. The equations used are given below.

Static Equilibrium conditions:

$\sum F_{x} = 0 \sum F_{y} = 0 \sum τ = 0$

(a) Calculation for the Fh

Using the given figure in the problem and applying static equilibrium conditions:
$\sum F_{x} = 0 F_{h} - F_{3} = 0 \cdot \cdot \cdot \cdot \cdot \cdot (1) \sum F_{y} = 0 F_{v} - F_{1} - F_{2} = 0 \cdot \cdot \cdot \cdot \cdot \cdot (2) \sum = 0 (F_{v} \times d) - (F_{2} \times b) - (F_{3} \times a) = 0 \cdot \cdot \cdot \cdot \cdot \cdot (3)$

From equation (1):

$F_{h} - 5 = 0$

Hence, $F_{h} - 5 = 0$

(b) Calculation for the Fv

From equation (2):

$F_{v} - 20 - 10 = 0$

Hence, $F_{v} = 30 N$

(c) Calculation for the d

From equation (3):

$(30 \times d) - (10 \times 3) - (5 \times 2) = 0 d = 1.3 m$

Hence, d = 1.3 m

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Short Answer

Step by step solution

Understanding the given information

Concept and formula used in the given question

(a) Calculation for the Fh

(b) Calculation for the Fv

(c) Calculation for the d

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