A box of mass 80kg is to be pulled along a horizontal floor...

A box of mass 80kg is to be pulled along a horizontal floor by means of a light rope. The rope is pulled with a force of 100N and the rope is inclined 20 degrees to the horizontal A. Calculate the frictional force between the box and the floor and also the normal reaction If the frictional force is 120N and the box is now pulled along the floor with the rope always inclined at 20 degrees to the horizontal B. Calculate the force with which the rope must be pulled for the box to move at a constant speed. C. Calculate the acceleration if the force Is 140N

Transcript

00:01 Hi guys, now in this column box of 80 keys is pulled along a horizontal flow by the means of light rope.

00:06 The rope is pulled with a force of 100 newton and the rope is inclined at the angle of 20 degree to horizontal a.

00:13 Now we need to calculate the frictional force between the box and floor and also we need to find out normal direction.

00:23 If the friction force is equal to 120 newton and the box is pulled along the floor with a rope or a rope or, always inclined at the angle of 20 degree to the horizontal b, calculate the force with which the rope must be pulled for the force to move at a constant speed, calculate the acceleration if force is 140 newton.

00:48 This is a box that we have and this box has a mass of what 80 kg.

00:54 A force applied by this angle is at 20 degree.

01:00 Let me draw a horizontal line we have.

01:02 This is 20 and this angle is 20 degree now this one is what this is 100 norton let's draw the components the component of 100 each what this one and again this one fbd of this of sight is what this is mg which is acting downward normal which is acting upboard just perpendicular to the contact surface and friction force will be worked so this is normal f of n here we have this is what m g this is 100 cos of 20 degree.

01:43 So this is what 100 cost 20.

01:47 This is 93 .97 newton.

01:59 And this is 100 sign of 20.

02:05 That is 34 .2 newton.

02:11 Mg, which is what? this is 80 times 9 .8.

02:15 That is 784 newton this is a friction force so first we need to calculate the first problem we need to calculate friction and normal reaction so normal reaction is equal to what i can say summation of force in y direction is equal to 0 which means i'm going to say that force this is normal direction f of n is equal to 784 minus 34 .2 and that is equal to what is 784 minus 3 4 .2 and that is equal to 4 .2 and that is equal to 749 .8 newton.

02:57 And similarly if this is not moving in forward direction i consider the frictional force sign let me say that the summation of fx is equal to 0 so frictional force is equal to 93 .97 newton.

03:14 Now if the frictional force is 120 newton and boxes pull along the floor with a rope always inclined at the angle of 20 degree...

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