Answer to Question #115556 in Mechanics | Relativity for isaac

Question

forces 5N, 12N and 13N acting on a body are in equilibrum. What is the angle between the forces 5N and 12N

Answer 1

Solution.

$F_1=5N;$

$F_2=12N;$

$F_3=13N;$

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0;$

$\overrightarrow{F_1}+\overrightarrow{F_2}=\overrightarrow{F_3};$

f the forces are perpendicular, then we find the equivalent of the two forces by Pythagoras' theorem, otherwise we will use the cosine theorem.

$\overrightarrow{F_{12}}=\overrightarrow{F_1}+\overrightarrow{F_2};$

First, check whether the forces are perpendicular:

$F_{12}=\sqrt{F_1^2+F_2^2};$

$F_{12}=\sqrt{(5N)^2+(12N)^2}=13N;$

$F_{12}=F_3;$

Therefore, the forces 5N and 12N form an angle of 90^o;

Answer: the forces 5N and 12N form an angle of 90^o.