The picture to the left shows a bucket connected to a...

The picture to the left shows a bucket connected to a massless rope, which runs over a massless, frictionless pulley. A man standing inside the bucket pulls the rope downwards in order to raise himself upwards. The mass of the man is 80.0 kg and the mass of the bucket is 20.0 kg . (a) How hard must the man pull the rope for him and the bucket to ascend at a constant speed? (b) Calculate the magnitude of the force that is needed for an upward acceleration of 1.2(m)/(s)2. (c) Using the man's free-body diagram, find the normal force exerted on the man by the bucket in parts (a) and (b). A block of mass m slides across a frictionless floor and then up a frictionless ramp. The angle of the ramp is \theta and the speed of the block before it starts up the ramp is v_(0) - The block will slide up to some maximum vertical height it above the floor before stopping. Derive an equation/expression for h in terms of m,\theta ,g and v_(0) Note: h may not depend of all of these quantities --- upon which of these quantities does h NOT depend?

Transcript

00:01 Hi there, here a box which has a mass of 80 kilograms is pulled by by force of 100 newtons along a horizontal floor by light rope which is inclined at 20 degree to the horizontal so first we have to explain why friction force must present let's illustrate this box this box is exerting the pulling force f and it's exerting a reaction normal reaction on the floor and gravity and let's look at this picture let's look at the scheme now if there is no friction force so if there is no friction then the system cannot be at equilibrium because fx will not be balanced and here f x is a horizontal projection of the force f on the x -axis so the green one is fx which equals to f times cosine alpha so therefore the friction force must be here and now let's so that's basically the answer to question a so now let's show the force diagram in question b and we've kind of done it already so we just need to add the friction force here now that's the force diagram and that's answered to question b now in question c we have to calculate the friction of force between the box and the floor the normal reaction force and the coefficient of friction let's start with a friction force equals to f times cosine alpha that is 100 newtons times cosine of 20 degree let's calculate it that is 93 .97 neutons which is roughly 94 newtons now let's we've calculated the friction force now let's calculate the normal reaction normal reaction equals to m g minus f times sine alpha which is 80 kilograms times 9 .8 meters per second squared minus hundred newtons times sign of 20 degree let's calculate it that equals to 750 newtons and now let's determine the coefficient of friction with k that equals to the friction force divided by n which is 94 neutrons divided by 750 newtons.

04:09 That equals to, okay, here we can round it, take it at 0 .125, because this friction forces, 94, can be written as 94 .0.

04:20 So, let's answer to question c.

04:24 And now let's move on.

04:27 In question d and e, the frictional force is increased to 120 newtons.

04:45 So let's rewrite, yeah, let's rewrite this task a little bit, so we have to clean up the space.

05:01 The xs are still the same, but now f is unknown, but the friction force fk is 120 newtons.

05:14 So in question d, we have to calculate the force f needed to move at a constant speed.

05:20 So therefore f x equals to the new friction force which is and fx also f times cosine alpha alpha is 20 degree therefore f equals to friction force divided by cosine of alpha which is 120 neutrons divided by cosine of 20 degree.

05:51 That is 127 .7 newton so we can round it to 128 neutrons.

05:58 So that's a answer to question d.

06:01 Now let's answer question e.

06:03 And here, the box is now pulled with a force of 140 newtons...

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