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Problem 49 Hard Difficulty

ssm A person is trying to judge whether a picture (mass 1.10 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.660. What is the minimum amount of pressing force that must be used?


$F_{N}=16.3 \mathrm{N}$


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Video Transcript

in this question. There are four forces acting on the picture. The weight force, the force that's being used to press it against the wall, the normal force exerted by the whoa as a response to the force that expressing the picture against the wall and the fish, no force that holds the picture in place. Then we have to calculate what is the minimum amount off that pressing force f to keep the picture in place. Then we have to use Newton's second law solve this question. We have to use it in two different access, so we have to use it. The vertical access on in these outside pointing access. Let me call these axes white, and these axes acts and let me choose it to be such that everything that is pointing outside the wall is positive and everything that it's pointing up his positive truth. Then we apply Newton's second law on each after taxes through for the vertical axis, we have the following The Net force acting on the vertical axis is it goes to the mass off the victor. Times acceleration off the pictures in the vertical axis. We want the picture, too, to be standing still. We wanted not to move, and it's not moving previously. Therefore, its acceleration is equal to zero. Then the net force in the Y direction must be equal to zero. But the net force on the white directions composed by two forces the original forest that is pointing up and therefore is positive in our reference free and the weight force, which is pointing downward. So it's negative reference free, and this is a question zero. So as a consequence, the frictional force must be equal to the weight force, and this is our first conclusion. So the frictional force must be equal to the weight force. But the frictional force is given by the following expression in the situation where the object isn't moving. It's too close to the static, which no coffee shelter times, the normal force that is acting on it, and this is equal to the weight force. Then we concluded that the normal force that is acting on the picture must be equal to the weight force divided by this tactic. Frictional confusion. No, we have playing Newton's second law on the other axis to get morning for mission, So the net force on that order. Access is equal to the mass times acceleration in that order direction again, We want to picture not to move in this direction to so acceleration is close to zero again. Then then that force on that direction is the question zero. But the net force in the direction is given by true forces. The force that is France in the picture against the wall and the normal forced. The normal force is pointing outside the wall. Therefore it's positive and the forced expressing the picture against the world is pointing to the inside. Then it's negative and these is equal to zero. Therefore, we got the relation between the normal force and the force that expressing the picture against the world. But we have already a relation between the normal force, the weight forced and the static frictional coefficient. And this is the normal force that must be acting on that picture to hold it in place. Therefore, the forced half that must be pressing the picture to keep it in place must be equals to the weight force divided by the static frictional coefficient. Then remember that the weight forces given by the Mass off the picture times acceleration of gravity, and this is divided by the static regional coefficient. Also remember that the acceleration of gravity near the surface off the earth is approximately 9.8 meters per second squared, then substituting the values that were given by the problem. We got 1.1 times 9.8 divided by 0.626 Is there close to the minimum amount of force that must be applied in the picture to hold it in place? And this gives us a minimal force off approximately 16 0.3 noodles, and this is the answer to the question.