Problem 4 inclined plane box of mass m 10 kg starts from...

Problem 4: Inclined plane box of mass m = 10 kg starts from rest on an incline and slides down to the bottom; & distance d = 10 m away: The angle of the incline is 30° and there is no friction. Use the 7-Step Recipe to Solve Problems with Forces to find the: 1) the acceleration of the block down along the incline; 2) the normal force acting on the box. Explicitly show in your solution each of the Steps. b) Using the acceleration that you found in the previous part, use the appropriate kinematics equations and initial conditions to find the time it takes for the box to get to the bottom of the incline.

Transcript

00:01 Here in the first part we need to represent the forces acting as seven step as mentioned in the question.

00:12 So this is how it will be.

00:14 In the first part you can see, this is step one, you can see the mass on an inclined surface.

00:20 In the next part, we have changed the mass into a point mass.

00:24 Now in the third step we have marked the angle along with the point mass.

00:31 This will be the point mass along with the inclined angle.

00:35 Now in the fourth step you can see the mg factor acting downwards.

00:40 Here the mg is resolute into two components.

00:44 Mg sine theta acting along the inclined surface and mg cos theta acting in this direction that is the opposite to the normal reaction n.

00:54 So this will be the fourth step.

00:57 Now we are moving on to the fifth step here.

01:00 In the fifth step, we will calculate that here f of x is equal to mass into acceleration a x.

01:11 Here f of x is equal to mg sine theta, that is sine 30, is equal to mass into ax.

01:22 Let this be equation 1.

01:23 Now f y will be equal to mass into acceleration in the y direction so here fy is equal to m g cos 30 will minus n that is m g cos 30 is acting in this direction but n factor is acting directly opposite that is normal reaction will be acting directly opposite so the total forces on y direction will be m g cos theta that is m g cos 30 minus n will be equal to mass into acceleration in the y direction so let this be equation number two now in the next step in the step six let us calculate the constraints so here constraints will be given that the according to the question given that the friction is equal to zero that is the first constraint and acceleration in the y direction will be equal to 0.

02:32 So in the step 7 we can say that from the constraints given we can say that from equation 1 we have from equation 1 we have mg sine 30 degree is equal to mass into acceleration in the x direction or acceleration in the x direction will be equal to here the mass get cancelled each other so g sine 30 value of g is equal to 9 .81 meter per second square into sine 30.

03:08 Upon solving this we will get an acceleration a x acceleration in the x direction to be 4 .9 meter per second square so this will be the acceleration so this is one answer now we are moving on to the next part from equation number 2 we have from equation number 2 we have that mg cos 30 minus n is equal to mass into acceleration in the y direction according to the question acceleration in y direction equal to 0 so m g cos 30 will be equal to the normal reaction n or we can say that the normal reaction n will be equal to mg cos 30 here mass is given 10 into acceleration due to gravity 9 .8 meter per second square into cost 30.

04:06 Upon solving this, we will get the normal reaction n to be 84 .87 newton.

04:15 So this will be the normal reaction...

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