Question

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???

Answer 1

you will use the equation for work which is w= F times distance so you will take 40N times 36N which equals 1440N

Answer 2

Answer:

The horizontal component of this force is about 36 N.

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 :

$F_x=F\ \cos\theta$

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

$F_x=35.95\ N$

or

$F_x=36\ N$

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