A 5.0-kg block rests on a 30^∘ incline. The coefficient of...

A $5.0-\mathrm{kg}$ block rests on a $30^{\circ}$ incline. The coefficient of static friction between the block and the incline is $0.20$. How large a horizontal force must push on the block if the block is to be on the verge of sliding $(a)$ up the incline and $(b)$ down the incline?

Key Concepts

Resolving a Horizontal Force on an Inclined Plane

When an external force is applied horizontally to an object on an incline, it must be decomposed into components parallel and perpendicular to the inclined surface. Understanding this transformation is critical because it affects both the net force that drives the motion up or down the incline and the normal force, which in turn influences the frictional force.

Equilibrium Conditions

Equilibrium conditions occur when the sum of forces acting on an object is zero, resulting in a state of no acceleration. In problems involving impending motion, setting up these conditions by equating force components allows one to solve for unknowns such as the magnitude of an applied force needed to overcome static friction.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It is proportional to the normal force through the coefficient of static friction, and its maximum value defines the threshold before motion occurs. This concept is crucial when determining conditions under which an object on an incline begins to move.

Force Decomposition on an Incline

This concept involves breaking down forces into components that are parallel and perpendicular to an inclined surface. It is essential for analyzing motion on inclined planes, as the gravitational force is resolved into these components to understand their individual effects on the object's propensity to slide or remain stationary.

Transcript

00:01 In this problem, we're asked to consider a 5 -kilogram rock sitting on a 30 -degree incline.

00:05 The coefficient of this situation is 0 .20.

00:09 We're asked how large a horizontal force must push on the block.

00:15 First, to be on the verge of sliding up the incline, and second, down the incline.

00:23 Okay, so let's figure out for first, up the incline.

00:39 So let's do it for some of our forces.

00:43 For a perpendicular has equal zero.

00:46 So rn are normal force minus mg times the cosine of 30 degrees.

00:57 Oh, let's not put my numbers in yet in case i can get rid of any of this will equal zero.

01:10 So that we'll know that rn equals mg cosine of theta plus f times the sign of theta.

01:26 Okay.

01:28 I was about to move upwards, my parallel.

01:34 Okay, this is.

01:45 And then, so this will be my net force parallel to the incline.

01:58 So my force parallel will be equal to zero, and that'll be f times the cosine of theta minus mg times the sign of theta minus the static friction has to equal zero.

02:16 My static friction will be equal to this coefficient of static friction times my normal force.

02:38 And this will give me coefficient sliding friction times m .g times the cosine of theta plus f times the sign of theta equals zero.

03:11 So now we can put things a little bit more conveniently.

03:38 Okay.

03:39 I'm going to move this up here a bit.

03:41 Take our fs out.

04:08 Still here.

04:12 So my force will now equal.

04:35 There we go.

04:37 Now we can plug in our vertical, we can plug in our values.

05:28 And that'll be divided by.

05:42 Okay.