Mary and her sister are playing with a cardboard box on the...

Mary and her sister are playing with a cardboard box on the neighborhood hill. Mary climbs into the box, and the total mass of the box with Mary in it is 115 kg. The box starts at rest at the beginning of the incline. The hill is at an incline of 28 degrees with respect to the horizontal. The static and kinetic friction between the box and hill is 0.4 and 0.2 respectively. Assume that Mary is now in the box but has not started to move. a. What is the magnitude of the friction force acting on the box? b. What is the magnitude of the net force acting on the box and child system? c. Will the box slide down the hill or stay at rest?

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 Here it's given that mary and her sister are playing with cardboard box on the neighborhood hill.

00:09 Mary climbs into the box and total mass of the box with mary in it is 115 kgs.

00:18 The box starts at rest at the beginning of the incline.

00:21 The hill is inclined at an angle of theta which is 28 degrees with respect to the horizontal like this.

00:31 This is angle theta.

00:35 The static and the kinetic frictions between the hill and the box is static is 0 .4 and kinetic is 0 .2 respectively.

00:43 Assume that the mary is now in the box and has not started to move.

00:48 Now you have to calculate the magnitude of the frictional force acting on the box.

00:53 Right.

00:54 So here you have to do two things.

00:57 So mg is acting downwards.

00:59 So here the component is mg sine theta that is acting along this incline and you have got mg cos theta perpendicular to and this is raised by the normal reaction.

01:11 You have got the frictional force acting right.

01:14 Now first you have to calculate these two parameters.

01:17 So mg sine theta gives you a value of 115 times g is 9 .8 sine of 28 degrees.

01:27 Right.

01:27 So this is equal to 529 .1 newton.

01:39 Right.

01:39 So this is mg sine theta.

01:40 And next you have to calculate the static coefficient of friction which is mu s times the normal reaction which is mu s times mg cos theta.

01:52 Right.

01:53 Mu s is 0 .4.

01:55 Mass is 115.

01:57 G is 9 .8 and here you have got cos of 28 degrees...