Bruce (the dog) is pulling on his lead with a force of 40N at an angle of 26º below the horizontal. Show that the horizontal component of this force is about 36N. Help???

Question

asked Jan 2, 2017 105k views

2 Answers

Johnny Westlake · Answer 1 · 2017-01-07T03:55:16+0000

Answer:

The horizontal component of this force is about 36 N.

Step-by-step explanation:

Given that,

Bruce is is pulling on his lead with a force of 40 N

Angle below horizontal,
$\theta=26^(\circ)$

We need to find the horizontal component of this force. The horizontal component of any vector is given by :

$F_x=F\ \cos\theta$

$F_x=40* \ \cos(26)$

$F_x=35.95\ N$

or

$F_x=36\ N$

So, the horizontal component of this force is about 36 N. Hence, this is the required solution.