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Problem 99 Hard Difficulty

ssm A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 $\mathrm{N}$ . The coefficient of static friction between his hands and the book is $0.40 .$ To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?


39 $\mathrm{N}$


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Video Transcript

in this question. By pressing the sides off the book, a frictional force is produced pointing upwards. We have to discover what is the minimum magnitude off the present force that is needed to keep the book in place. In order to answer that question, have to use Newton's second law are you? Choose the following reference frame a vertical axis pointing upwards and the horizontal axis pointing to the right, then applying. Newton's second law, the vertical axis, results in the following the Net Force. The Y direction is given by the mass off the book times its acceleration in the direction it should not be moving say its acceleration is constrained to be close to zero and then that force is a question zero. But the net force in the Y direction is composed by two forces the frictional force which points to the positive Y direction the weight force which points to the negative y direction. So fiction off warrants miners. The weight forced is it goes to zero before different No force is he goes to the weight force. Okay, this is a nice result. No, that is moved to the X direction in order to get something with the force f. So in the horizontal direction we have the following the net force on that direction is there close to the mass off the book times. It's horizontal acceleration, which is a question zero because the book should stay in place. Then there are two forces acting on the horizontal and they are equal forces acting on opposite directions before we got to zero equals 202 These applications off Newton's second law on the horizontal direction have not helped us. Then what we can do? You can remember that if the force f is the minimum force for the book not to fall, then the frictional force should be the biggest possible frictional force, which in the case that the books not moving is a static frictional force given by the static snow coefficients times the normal force. But then, what is the normal force that is acting on the book? In the situation, the force F is behaving as the normal force on. There are two forces that are pressing, so we have the normal force that is equals two times F because the total pressing force is the close to true times f. So that's it, then these give us the following equation. The static original coefficients times half times true is it goes with the weight force. Then the necessary force is given by the weight forced, divided by two times less static friction or proficient. Then we have to use the data that the problem gives us, which is that the weight force is 31 new terms, and the static frictional coefficient is 0.4. Then the pressing force is given by 31. Divided by two times is Europe 14 which is 31 divided by 0.8. And these results in a Princeton force off approximately 39 new tones. And this is the answer to these questions.

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