Question

While sliding a couch across a floor, Hannah and Andrea exerts forces FH and FA on the couch. Hannah's force is due north with a magnitude of FH = 59 N and Andrea's force is θ = 22° east of north with a magnitude of FA = 155 N. In this problem, use a coordinate system with y directed north and x directed east.
Part (a) Find the net force in the y-direction in Newtons.
Part (b) Find the net force in the x-direction in Newtons.
Part (c) Calculate the angle in degrees north of east of the net force exerted on the couch by Hannah and Andrea, F HAF→HA.
Part (d) Hannah and Andrea's housemates, David and Stephanie disagree with the move and want to prevent its relocation. Their combined force F DSF→DS must be equal and opposite to that of FHAF→HA. What is the magnitude in Newtons of the force F DSF→DS which will prevent the relocation?

Answer 1

Answer:

a) 202.7 N

b) 58.1 N

c) 74.1º N of E.

d) 210.9 N

Explanation:

a)

• The net force exerted in the y-direction, will be the sum of FH (which is directed northwards) and the y-component of FA.
• Since the magnitude of FA is 155 N and the angle of FA with the y-axis, is 22º (E of N), we can find the N-S component of  FA, just applying the the definition of cosine, to the triangle defined by FA, the y- axis and a segment parallel to the x- axis between FA and the y-axis, as follows:

      $F_{Ay} = F_{A} * cos \theta = 155 N* cos 22 = 143.7 N (1)$

      ⇒ Fy = FH + FAy = 59 N + 143.7 N = 202.7 N (2)

b)

• We can proceed exactly in the same way for the x-axis.
• Since FH is directed due North, it has no component along the x-axis.
• So, Fx is directly the component of FA along the x-axis, which can be found applying the definition of sine to the same triangle than in a) as follows:

      $F_{x} = F_{A} * sin \theta = 155 N* sin 22 = 58.1 N (3)$

c)

• Taking the same triangle than in a) and b), we can apply the definition of tangent, in order to find the angle between F and the x-axis, as follows:

       $tg \theta = \frac{F_{y}}{F_{x}} = \frac{202.7N}{58.1N} = 3.5 (4)$

      ⇒ θ = tg⁻¹ (3.5) = 74.1º N of E. (5)

d)

• In order to be equal and opposite to the combined force FH+FA, it must have the same magnitude.
• This magnitude can be found applying the Pythagorean Theorem to the same triangle that we used in a), b) and c):

      $F_{DS} = \sqrt{(F_{x} ^{2} +F_{y} ^{2})} = \sqrt{(58.1N)^{2}) + (202.7N)^{2} } = 210.9 N (6)$