Question

question 4.1c
the block shown in the figure is sliding down a frictionless surface inclined at an angle θ = 33.4° above the horizontal. the blocks mass is m = 4.55 kg.
determine the magnitude of the normal force n acting on the block.

Explanation:

Step1: Analyze forces on block

The gravitational force on the block is $F_g = mg$, where $m = 4.55$ kg and $g=9.8$ m/s². The normal force $n$ is perpendicular to the inclined - plane. Resolve the gravitational force into components. The component of the gravitational force perpendicular to the inclined plane is responsible for the normal force.
$F_g=mg$
$F_g = 4.55\times9.8=44.59$ N

Step2: Find normal - force formula

The normal force $n$ on the block on an inclined plane is given by $n = mg\cos\theta$, where $\theta = 33.4^{\circ}$.
$n=F_g\cos\theta$

Step3: Calculate normal force

$n = 44.59\times\cos(33.4^{\circ})$
$n = 44.59\times0.8356$
$n\approx37.2$ N

Answer:

$37.2$ N