a 217 kg block is held in equilibrium on an incline of angle q 596o by a horizontal force f appl

First, we need to calculate the forces acting on the block. There are three forces acting on the

A 217 kg block is held in equilibrium on an incline of angle...

A 2.17 kg block is held in equilibrium on an incline of angle q = 59.6° by a horizontal force, F, applied in the direction shown in the figure below. A) If the coefficient of static friction between the block and incline is μs = 0.308, determine the minimum value of F. B) Determine the normal force of the incline on the block.

Transcript

00:01 In this problem, we have a block in equilibrium on a inclined plane that makes an angle and stata with the resultant.

00:22 And there is a horizontal force applied on d.

00:28 So let us draw all the forces acting on the block.

00:32 Since the problem wants us to turn the magnitude of the applied force and the normal force, given that the mass of the block is 2 .17 kilograms and the angle of inclination of the inclined plane is 59 .6 degrees.

00:53 So the force is acting on the block is this.

00:55 Suppose this is the center of mass of the block, vertically downwards is the weight.

01:02 Let's denote as f sub -w.

01:04 And this is the applied horizontal force.

01:10 Perpendicular to the surface of the inclined plane is the normal force.

01:16 Sub -n and opposite the normal force is the projection of weight in this direction, which is equals to weight multiplied by cosine theta, since this angle, which is by geometer is then the projection of the applied force in this plane that's not as y -plane so -so -fane so in this direction, we have the friction force, static friction force, and opposite it is the projection of the weight in this direction.

02:12 So as x -plane, which by trigonomy is then f -w sine theta.

02:21 Take note that this angle is also theta.

02:24 Then the projection of the applied force in this plane is f -a -posin theta, while on this direction is f -a, then writing down the force equation by first law of motion for abiding equilibrium the net force is zero so let's start with the y plane and the net force is the sum of the forces acting on the plane which are the normal force which is positive the projection of the applied force in this plane which is negative since opposite the normal force as well as the weight projection of weight or his plane shows to negative since opposite the normal force then isolating the normal force to the left side, then our equation becomes like this...

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