Problem 1: Determine the minimum horizontal force P required to hold the crate from sliding down

Step 1: Identify the forces acting on the crate. These include the weight of the crate (W = mg),

Problem 1: Determine the minimum horizontal force P required...

Transcript

00:01 Where in this question we're given that we have a crate whose mass is 50 gage on this crate there is a force acting which is p and the coefficient of static friction between the plane and the crate is given as 0 .25 we have to find out the minimum value of the force p so that the crate does not slide down let us first see what are the force that is acting on the crate first you'll have the gravitational force which always x downward and the gravitational force is given by the mass of the object m multiplied by x relation to gravity g then we'll have a normal force or reaction force the normal or the reaction force x perpendicular to the surface so we have our normal force you the friction force will act up the incline since the crate has a tendency to come down and we take the friction force to be f the gravitational force m g will have two components it will have a vertical component which is m g cost 30 degree and it'll have a horizontal component the horizontal component will be down the inclined plane and it will be m g sine 30 degree this is the angle between the gravitational force m g and its component which is perpendicular to the plane is 30 degree the force p will also have two components one along the plane which is p cause 30 degree and the other one which is perpendicular to the plane which will be p sign 30 degree.

01:36 Now to further solve this question, we'll apply newton's second law, according to which the sum of all the forces in a particular direction is equal to the mass of the object multiplied by its acceleration.

01:53 In our case, the crate is not moving, that means the acceleration will be equal to 0.

01:58 First of all, we'll apply newton's second law along the plane.

02:02 Along the plane we have p cost 30 degree plus the frictional force f plus negative of m g sign 30 degree equal to 0 because the acceleration is 0 we've taken negative sign here for mg sign 30 degree because it is acting down the plane while the other two forces are acting up the plane this gives us p cause 30 degree plus the force of friction which is given by the coefficient of static friction multiplied by the normal force n minus m g sign 30 degree is equal to 0 to mark this as equation number 1 now considering the direction which is perpendicular to the plane and applying newton's second law we get p sign 30 degree plus m g cost 30 degree minus n is equal to 0 we've taken negative sign for the normal course because it is acting in the upward direction while the other two forces are acting in the downward direction from here we will get the normal force n is equal to p sine 30 degree plus m g cos 30 degree we mark this as equation number 2 so using equation number 2 in equation number 1 we will get p cos 30 degree plus the coefficient of static friction multiplied by p sine 30 degree plus m g coss 30 degree minus m g sine 30 degree is equal to 0 we can observe that we have p in two different terms so let us take p as common term we'll get p multiplied by cos 30 degree plus the coefficient of static friction multiplied by sine 30 degree then we have mj as the common term here so plus m g multiplied by the coefficient of static friction multiplied by cos 30 degree minus sine 30 degree is equal to 0 we now put the known values so p multiplied by cost 30 degrees under root 3 divided by 2 plus the coefficient of static friction is 0 .25 multiplied by sine 30 degrees 1 divided by 2 plus the mass of the object is 50 kg multiplied by the acceleration into gravity g is 9 .8 meter per second square.

04:46 Multiplyed by the coefficient of static friction which is 0 .25 multiplied by cost 30 degree which is root 3 divided by 2 minus 10 30 degree which is 1 divided by 2 equal to 0...

Question

Problem 1: Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is $mu_s = 0.25$.

Question

Problem 1: Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is $mu_s = 0.25$.

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