Question

Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R. What is the magnitude of R to the nearest Ib. P = 568 lb T = 687 lb ---- - *-- - C/60° DI

Answer 1

Solution :

Given Data :-

P = 568 lb

T = 687 lb

First, we will find angle (a).

From ABCD the length of CD and BD is given as:

cos(60) = CD + CD= 3.......... (1

BD sin(60 sin(60) = - → BD = 5.1962'.............

From ABAD we will get angle a :

:. tan(a) = BD BD ADAC + CD 5.1962 3+3 3 = 0.866025

$\therefore \, \, \alpha =40.893^{\circ}\, \, \, .....................(3)$

Applying Law of triangle for vector addition of T and P:

The triangle for vector addition is shown below -

$\textup{Using Law of Cosine for above triangle : }$

$\therefore R^2=P^2+T^2-2(P)(T)\cos(\alpha )$

$\therefore R^2=(568)^2+(687)^2-2(568)(687)\cos(40.893 )$

$\therefore R^2=204638.3417$

${\color{DarkBlue} \therefore R=452.37\, \, \textup{lb}}\, \, \textup{(Answer)}$

$\textbf{Extra : }$

$\textup{Using law of sines : }$

$\therefore \frac{T}{\sin(\theta )}=\frac{R}{\sin(\alpha )}$

$\therefore \frac{687}{\sin(\theta )}=\frac{452.37}{\sin(40.893 )}$

...0 = 83.823"