A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined

A 68 kg crate is dragged across a floor by pulling on a rope...

Key Concepts

Equilibrium Conditions

Equilibrium conditions occur when the sum of all forces acting on an object in any direction is zero. In problems of impending motion, setting up equilibrium equations allows one to solve for unknown forces, such as the minimum force needed to overcome static friction or adjusting the normal force when additional forces are applied at an angle.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two contacting surfaces. It acts in response to an applied force up to a maximum value that is proportional to the normal force. Understanding static friction is essential when determining the threshold force required to start moving an object at rest.

Kinetic Friction

Kinetic friction is the resistive force acting between surfaces already in relative motion. It is typically lower than the static friction and is proportional to the normal force via a kinetic friction coefficient. This concept is crucial in calculating the acceleration of a moving object once it begins to slide.

Force Decomposition

Force decomposition involves breaking a force vector into its horizontal and vertical components. This is particularly important in problems where forces are applied at an angle, as it helps in accurately determining the contributions of the force in each direction, which subsequently affect the net force and the normal force.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object. It plays a key role in friction problems since both static and kinetic friction are directly proportional to this force. Adjustments in the normal force, due to additional vertical components of applied forces, directly influence the amount of friction present.

Newton’s Second Law

Newton’s Second Law of Motion states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle is used to relate the forces acting on an object, including friction and any applied forces, to determine its acceleration.

Transcript

00:01 Free body diagram for the mass, going straight up would of course be force normal.

00:07 Going straight down would be mg, gravitational force.

00:12 At an angle here defined theta, we have the tension from the rope, and then opposite to the direction of motion, be frictional force.

00:27 After drawing the free body diagram, we can then apply the newton's second law in the x and y directions.

00:33 So t, the tension cosine of theta minus the frictional force with be equaling zero.

00:40 This is the system at a constant velocity or of course at rest.

00:45 And we can say that t sign of theta plus the force normal minus the gravitational force m .g.

00:56 This will equal zero.

00:57 And we know that theta in this case is equaling 15 degrees.

01:01 We can say that here the first equation gives that the frictional force would be equaling to t cosine of theta and the second gives that the force normal is going to be equaling to the gravitational force mg minus t sine of theta and we know that if the crate is to remain at rest this means that the frictional force must be less than the coefficient of static friction multiplied by the force normal or we can say that t cosine of theta must be less than the coefficient of static friction and then substituting it for the force normal mg minus t sine of theta.

01:50 We can solve for, we can say when the tension force is sufficient to just start the crate moving, so just in order to start the crate removing, just exceeding that maximum static frictional force, we can say that t cosine of theta will be equalling to the coefficient of static friction mg minus t sine of theta.

02:19 And for part a, we can solve for the tension.

02:23 The tension t would then be equalling the coefficient of static friction times the gravitational force divided by cosine of theta plus the coefficient of static friction multiplied by sine of theta.

02:36 We can solve, this would be 0 .50, multiplied by 68 kilograms, multiplied by 9 .8 meters per second squared...

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