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$\bullet$$\bullet$ You push with a horl-zontal fonce of 50 $\mathrm{N}$against a 20 $\mathrm{N}$ box, press-ing it against a rough ver-tical wall to hold it in place, The coefficients of kinetic andstatic friction between this box and the wall are 0.20 and 0.50 ,respectively. (a) Make a free-body diagram of this box.(b) What is the friction force on the box? (c) How hard wouldyou have to press for the box to slide downward with a uni-form speed of 10.5 $\mathrm{cm} / \mathrm{s}$ ?

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

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Rutgers, The State University of New Jersey

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question to you. Three states that you push a box with a horizontal force of 50 Newtons against a 20 Newton box. Pressing eating is a rough vertical wall to hold it in place. The coefficients of kinetics and static friction Between this box and the wall are 0.2 and 0.5 respectively. Part of the question asks to make a free body diagram of this box. Preppy was a friction force on the box and seeing how hard we get the press for the boxes slide downward with a uniform speed of 10.5 meters per second centimetres per second. So you are aiming to try free body diagram. So this is our wall. We're holding our box against it. So we're and wanted to find my positive axes for X and y to be wide up next to the right, replying in force in this direction with locals, call F if our normal force is the victor across from the surface of the object, the interaction between the object of the surface session say, um, gravity or the wheat will put down what should be the direction motion. So we have the friction for us. I was generically called FF force A friction, um, pushing it up to prevent that motion downwards. That's part neighbor question party Bonified the friction force on the box. So, assuming it's static, which I believe the initially it is static so you can see that the net force in our direction has to be zero. And looking at her diagram, you see that the force, because it's not moving to get with stack Friction Force minus her weight are the two forces of the Y direction which have to equal zero instantly. You could say that our static friction forces simply quality to the weight, which is just 20. Ines, me and Percy, how hard would you have to press for the box? That's like downward with the uniform speed of 10.5 centimeters per 2nd 10.53 Um, one thing we don't really need to know the velocity per se if only she knows that the object will be moving. So in this case, we're gonna do with, um, um kinetic frictional force, not static. So if you look at first art net forces in the X direction, that, of course, again has to equal zero has a place in our diagram. Um, your force applied to the rights minus the normal force equals zero so that our forces just equipped to our normal for us. So that this question we're trying to solve with this force values So this term here is our unknown value. Now remove the underlying servicing confused with the letter e. In part, the rest discretion tells us that we're moving. We're not accelerating. So the net force would be zero as well in the Y direction. It's a constant speed. So that this time we have the force of kinetic friction has to do equal our weight as we found in part B of this question. And we can define that the force of connect friction is just being UK times the normal force supposed to wait, but we've got previously appear that are normal. First is just equivalent to the applied force. So we can rearrange this equation playing f for a normal for us. And we just have I couldn't skip a step tonight that confused even me. So let's not do that. We have our co vision of contrition. Times are normal, normal course, which again is just the same of the applied for us, would you could live to the weight so we can find that applied force. It's simply our weight divided by a coefficient of connect friction. The way it has given previously was 20 Newtons and a court vision correction is point to Oh, and by solving this, we would simply find that's a net force that needs to be applied is 100 millions and there you go.

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