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$\bullet$$\bullet$ Block $A$ in Figure 5.64weighs 60.0 $\mathrm{N}$ . The coef-ficient of static friction be-tween the block and thesurface on which it restsis $0.25 .$ The weight $w$ is12.0 $\mathrm{N}$ and the system re-mains at rest. (a) Find thefriction force exerted onblock $A .$ (b) Find the max-imum weight w for whichthe system will remain at rest.

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a. $F_{f}=w=12 \mathrm{N}$b. $w_{\max }=15 \mathrm{N}$

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

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

McMaster University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

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Block $A$ in $\textbf{Fig.…

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Block $A$ in Fig. P5.72 we…

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Block $A$ in Figure 5.70 w…

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Block A in Fig. P5.72 weig…

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(a) Block $A$ in Fig. 5.63…

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5.72 Block A in Fig: PS.72…

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Block B in Fig. 6-31 weigh…

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A 2.00-kg block is placed …

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A 2.5 $\mathrm{kg}$ block …

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A 25.0 -kg block is initia…

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Block $B$ in Fig. $6-31$ w…

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$\bullet$$\bullet$ Block $…

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A $2.00-\mathrm{kg}$ block…

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A 2.5 kg block is initiall…

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A $2.5 \mathrm{~kg}$ block…

States that block a In the figure shown here, the block on top of the table we have 60 Newton's, the coefficient of static friction between the block and the surface, on which arrests is 600.25 the way W hanging blow is 12 Newtons, and the system remains at rest. Part of this question asks to find the friction force exerted on Block A and B, the maximum weight for which the system will remain at rest. Okay, so this question is best suited, as most are white label is is little w sorry about that. We should start a free body diagram for each of these boxes that were dealing with. So for mass A on the table, of course, there'll be some weight associate with it, which I'm calling here to be W a weight of block A because it's a rope attached to the right hand side. We have the tension for us acting in that direction, and so if this system were to move, block a would move to the right so the friction force would act opposite of that motion. So it's a mess, and of course, there's a normal force associated with block A is well and first thing I should have done. Probably was also drawn by a set of axes and came back with my directions. And now for the second mass that we're hanging mass, we have w So it's weight of Cibolo is W. And of course, is the rope pulling up so attention for us in the room. Okay, so you want to find the coefficients that I'm sorry, Not the corporation. Static friction, um, deficient force of a FS depleting our forces in the X direction for block A well like this for both system. But the only forces acting on the system are the ones before block A in the X direction. Nasty was zero. Because in this kids, once it starts moving than the block starts to move. So then therefore there's no acceleration. So we find that our attention for us is positive to the right and a supposes by our static friction force fs that equals zero. What should just be in the tension will be equal in two fs, so stop particular particularly useful because you don't know attention. However, if look at our second mass here, the weight w that one, of course, has this net force of the wind Rex to be zero because it's not moving, which means that attention would equal the weight of the little mass. Therefore, if I combine these two equations, I confined that the weight is according to the static friction for us, which is given in the problem to be 12 minutes is so answered. Party. The friction for us is 12 Mutants party. We want to find the maximum weight for which the system will remain at rest. So basically, when yourself for the map the weight in which this condition it's just just exceeded again. We have detention the equipment to the definition of the static friction, u. N and by looking at the definition of her nothing after sorry, the balance of forces over here a safe F y for this figure, block A. That means that the tension must equal the weight of block A. So the normal force must equal the weight of block A. Where's that? I can use the substitute in for the crazy for attention, and so simply via us times away today, each of these terms are known coefficients. That friction told me 0.25 and the weight of blockades 60 Newtons. And so we can find the net attention which corresponds to the weight, of course. So I put engineer. But again, based on what we've found for the other part replaces with wheat is abortion, which is solving for the maximum weight that is able to be placed. Hanging over the edge of the table would be 15 Newtons, which again asked. But it does match attention so anything more than 15 minutes will cause the acceleration of the blocks. There you go.

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