Question

Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N).
A) What is the magnitude of the component of Tom's weight parallel to the ladder?
B) What is the magnitude of the component of Tom's weight perpendicular to the ladder?

Answer 1

Answer: A) $P_{x}$ = 564.4 N

              B) $P_{y}$ = 374 N

Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:

$hypotenuse^{2} = side^{2} + side^{2}$

$side = \sqrt{hypotenuse^{2} - side^{2}}$

side = $\sqrt{3^{2} - 2.5^{2}}$

side = 1.65

Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ($P_{x}$) and othe that is perpendicular ($P_{y}$). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.

The image in the attachment illustrates the described above.

A) $P_{x}$ = P sen θ

$P_{x} = P.\frac{oppositeside}{hypotenuse}$

$P_{x}$ = 680.$\frac{1.65}{3}$

$P_{x}$ = 564.4 N

B) $P_{y}$ = P cos θ

$P_{y} = P.\frac{adjacentside}{hypotenuse}$

$P_{y}$ = 680. $\frac{1.65}{3}$

$P_{y}$ = 374 N