Question

"The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body?"

Answer 1

Answer:

The answer is $F= 2.881$ pounds

Explanation:

We will start this question by representing the force vectors in Cartesian coordinate system:

• Force 1 acting on the right $F_{1} = (2, 0)$
• Force 2 acting upward $F_{2} = (0,5)$
• Force 3 acting at 45° $F_{3} = (3cos(45),3sin(45))$

Let the single force we need be represented by $F=(x,y)$

Since, the body is in equilibrium the summation of all the forces acting on it should be zero:

∑$F_{x} = 0$ (Sum of all forces in $x$ direction is equal to zero)

$x + 2 - 3cos(45) =0$ ($x$ component of force 3 is negative because it is acting  in opposite direction)

$x = 0.1213$

∑$F_{y}=0$ (Sum of all forces in $y$ direction is equal to zero)

$y+5-3sin(45)=0$ ($y$ component of force 3 is negative because it is acting  in opposite direction)

$y=-2.8786$

So,

$F=(0.1213,-2.8786)$

Hence, to get the scalar quantity of a single force:

$F=\sqrt[2]{(0.1213)^{2}+(-2.8786)^{2} }$

$F= 2.881$ pounds