A kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15^

A kg crate is dragged across a floor by pulling on a rope...

Key Concepts

Force Decomposition

Force decomposition is the process of breaking a force vector into its horizontal and vertical components, which is essential in analyzing the effects of forces applied at an angle. By resolving the force into orthogonal components, one can independently consider the forces acting in each direction, such as overcoming friction in the horizontal while affecting the normal force in the vertical.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It acts parallel to the contact surface and adjusts up to a maximum value determined by the product of the coefficient of static friction and the normal force. Understanding static friction is crucial to determining the minimum force required to start an object moving.

Kinetic Friction

Kinetic friction is the resistive force encountered when an object is already in motion relative to the surface it is sliding on. It is generally lower than static friction and is calculated as the product of the coefficient of kinetic friction and the normal force. This concept is important for analyzing the motion of objects once they have overcome static friction and are accelerating.

Normal Force Adjustment

The normal force is the force perpendicular to the contact surface that typically balances the weight of the object. When an additional vertical component is applied via an angled force, it alters the effective normal force. This adjustment is crucial because it directly affects the frictional forces, as friction is proportional to the normal force.

Newton's Second Law

Newton's second law states that the net force acting on an object equals the mass of the object multiplied by its acceleration (F = ma). This principle is fundamental for linking the forces acting on an object, such as applied forces and friction, to the resulting acceleration. It is applied separately in different directions after resolving forces, making it a key tool in solving dynamics problems.

Transcript

00:01 In this question, we have a 68 kilogram crate being dragged across the floor.

00:07 So let's first draw our free body diagram.

00:12 So we got force of gravity straight down, normal force up, force of friction, and our force that we're pulling with at an angle of 15 degrees.

00:28 Now we want to redraw this free body diagram, breaking that force at 15 degrees up into two parts.

00:34 So that means that upwards, it's going to be fn plus.

00:38 F sign of 15 degrees.

00:42 To the right is f cos 15 degrees.

00:47 Down is fg and force of friction.

00:54 And a couple other things that we know.

00:57 We know force of friction is equal to, or force of static friction is less than equal to mu -s times normal force.

01:09 Force of kinetic friction is equal to mu -k times normal force and fg equals mg.

01:23 From our food body diagram, we can say the following two things.

01:27 Up and down are going to be balanced, so fg is equal to f normal plus f sign 15 degrees.

01:40 And for part a, we want to know if the coefficient of static friction is 0 .5, what force is required to start the crate from moving? or in other words, we're looking for what point does the force of static friction equal the pulling force of f cosine 15? so let's move some stuff around.

02:02 In our top equation, we could rewrite this as fn equals fg minus f sign 15 degrees.

02:14 So now if we take our expression here for force of static friction, okay, and we're just going to use the equal to part.

02:20 So let's make this an equal to sign.

02:23 If we don't care about the last and they're equal to, we're looking for the tipping point, so we're looking for the equal.

02:30 So using my underlined green expression and then these two equations right here, what i can say is the following.

02:39 Mu s times fg minus sine 15 degrees equals f cosine of 15 degrees.

03:32 So if we do a little bit of algebra and rearrange here and make some substitutions, what we end up with is the following...

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