A 100 N force, directed at an angle θabove a horizontal floor, is applied to a 25.0 kg chair sit

A 100 N force, directed at an angle θabove a horizontal...

A 100 $\mathrm{N}$ force, directed at an angle $\theta$ above a horizontal floor, is applied to a 25.0 kg chair sitting on the floor. If $\theta=0^{\circ}$ , what are (a) the horizontal component $F_{h}$ of the applied force and (b) the magnitude $F_{N}$ of the normal force of the floor on the chair? If $\theta=30.0^{\circ},$ what are $(\mathrm{c}) F_{h}$ and $(\mathrm{d}) F_{N} ?$ If $\theta=60.0^{\circ},$ what are $(\mathrm{e}) F_{h}$ and (f) $F_{N} ?$ Now assume that the coefficient of static friction between chair and floor is $0.420 .$ Does the chair slide or remain at rest if $\theta$ is $(g) 0^{\circ},\left($ h) $30.0^{\circ},$ and $(i) 60.0^{\circ} ?\right.$

A 100 $\mathrm{N}$ force, directed at an angle $\theta$ above a horizontal
floor, is applied to a 25.0 kg chair sitting on the floor. If $\theta=0^{\circ}$ , what
are (a) the horizontal component $F_{h}$ of the applied force and
(b) the magnitude $F_{N}$ of the normal force of the floor on the chair?
If $\theta=30.0^{\circ},$ what are $(\mathrm{c}) F_{h}$ and $(\mathrm{d}) F_{N} ?$ If $\theta=60.0^{\circ},$ what are $(\mathrm{e}) F_{h}$
and (f) $F_{N} ?$ Now assume that the coefficient of static friction between chair and floor is $0.420 .$ Does the chair slide or remain at rest
if $\theta$ is $(g) 0^{\circ},\left($ h) $30.0^{\circ},$ and $(i) 60.0^{\circ} ?\right.$

Key Concepts

Force Decomposition

This concept involves resolving a force vector into its horizontal and vertical components using trigonometric functions such as cosine for the horizontal component and sine for the vertical component. It is crucial for analyzing the effects of a force applied at an angle relative to a chosen coordinate system.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object. When additional vertical forces are present, such as the vertical component of an applied force, the normal force is adjusted to maintain equilibrium by counteracting both the object's weight and the vertical forces acting on it.

Static Friction

Static friction is the resistive force that prevents two contacting surfaces from sliding relative to each other. Its maximum value, which is determined by the product of the coefficient of static friction and the normal force, is critical in assessing whether an applied horizontal force is sufficient to cause an object to begin moving.

Gravitational Force

Gravitational force, often referred to as the weight of an object, is the force exerted by gravity and is calculated as the product of mass and the acceleration due to gravity. This force affects the equilibrium conditions, particularly the determination of the normal force when other vertical forces are in play.

Transcript

00:01 So here we're going to say that the plus direction is horizontal to the right.

00:14 And then we can say a plus y is vertical up.

00:20 We know that the magnitude of the applied force is 100 newtons.

00:27 And so we can say that the x component of this force, which in the textbook is denoted as f sub h.

00:35 S sub h simply means the horizontal force.

00:38 So h just stands for horizontal.

00:40 I would say it's better to denote it as f sub x because x is referring to the cartesian axes and not simply just horizontal or vertical.

00:52 So this would be equal to f the magnitude of the force times cosine of theta.

00:58 And we know that this is going to be 100 newtons because the entire force in this case would be in the x direction.

01:11 So here we can say that there is no vertical acceleration.

01:16 So we can apply newton's second law for part b and say that fn plus force in the y direction would equal mg.

01:35 And we can then say force normal would equal mg minus the force sine of theta.

01:43 However, here we have a mass of 25 kilograms.

01:49 So we can actually just multiply 25 kilograms times the acceleration due to gravity.

02:04 And we find that the force normal in part b is 245 newtons given that theta equals zero degrees.

02:13 If theta equals zero degrees, we can eliminate this term.

02:17 And of course, again, the force normal would simply equal the weight.

02:21 However, for part c now, we have an angle of 30 degrees.

02:28 So when we have an angle of 30 degrees, now we actually do have to account for the angle.

02:34 And so the x component of the force, or we can say again, f horizontal, would be the magnitude of that force, cosine of theta.

02:46 We can say that this is going to be equal to 100 newtons times cosine of 30 degrees.

02:54 And this is giving us 86 .6 newtons.

02:59 For part d, again, the angle is 30 degrees.

03:03 And now we're trying to solve for the force normal.

03:06 We can use the exact same formula.

03:10 And we can say that the force normal would then equal mg.

03:15 Here it would be minus the magnitude of the applied force times sine of 30 degrees.

03:20 So this would be 25 kilograms multiplied by 9...

Please give Ace some feedback