Question

part d: vector addition using components

1. calculate the total displacement in the horizontal (x) and vertical (y) directions using the algebraic method and grasp. show all work in table 4.

table 4.

	δd_tx	δd_ty
	d_co = 27.6 m e	δd_de = 26.3 m s
r:	δd_tx =?	r: δd_ty =?
a:	δd_tx = δd_ab + δd_co	a: δd_ty = δd_bc + δd_de
s:	δd_tx = δd_ab + δd_co	s: δd_ty = δd_bc + δd_de
	δd_tx = 11.4 m e + 27.6 m e	δd_ty = 8.56 m s + 26.3 s
	δd_tx = 39.0 m e	δd_ty = 34.86 m s
		δd_ty ≈ 34.9 m s
p:	the total horizontal displacement is 39.0 m e	p: the total vertical displacement is 34.9 m s

Explanation:

Step1: Identify horizontal displacements

We have $\Delta\vec{d}_{AB}=11.4m[E]$ and $\Delta\vec{d}_{CD}=27.6m[E]$. The formula for total horizontal displacement $\Delta\vec{d}_{Tx}$ is $\Delta\vec{d}_{Tx}=\Delta\vec{d}_{AB}+\Delta\vec{d}_{CD}$.

Step2: Calculate horizontal displacement

$\Delta\vec{d}_{Tx}=11.4m + 27.6m=39.0m[E]$.

Step3: Identify vertical displacements

We have $\Delta\vec{d}_{BC}=8.56m[S]$ and $\Delta\vec{d}_{DE}=26.3m[S]$. The formula for total vertical displacement $\Delta\vec{d}_{Ty}$ is $\Delta\vec{d}_{Ty}=\Delta\vec{d}_{BC}+\Delta\vec{d}_{DE}$.

Step4: Calculate vertical displacement

$\Delta\vec{d}_{Ty}=8.56m+26.3m = 34.86m\approx34.9m[S]$.

Answer:

The total horizontal displacement is $39.0m[E]$ and the total vertical displacement is $34.9m[S]$.