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All right.
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Chapter 6, question 8 states, to move a large crate across a rough floor, you push on it with a force f at an angle of 21 degrees below the horizontal, as shown in the figure here.
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We want to be able to find the force necessary to start the crate moving, given that the mass of the crate is 32 kilograms and the coefficient of static friction between the crate and the floor is 0 .57.
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So i've drawn this situation here.
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We're trying to solve for the force as our unknown variable or unknown here.
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We have the mass of the crate or the angle of which the force is being applied, and a coefficient of static friction mu .s of 0 .57.
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So to start, we will, as we often like to do, well, first i will create an axis just for definition of x and y, which is always the useful first step.
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And we should apply a force body diagram to our box just so we can see all the forces being applied so of course we have force due to gravity which i'll call fg the applied force is at an angle shown here i'll call it f i'll just call it f that's what's given in the question our force applied at the angle of 21 degrees below the horizontal our force of friction force of static friction is opposing our direction of motion so f s and of course with every object we have opposing force due to gravity, obviously known as the normal.
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So we'll look at our forces in terms of their principal directional components.
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So we'll start with our forces in our x direction.
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So our applied force in this direction, if this is our angle here, separating it, the component of x due to the applied force is given by cosine.
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So it's f times cosine theta minus our frictional force, our force of friction is given by mu s times the normal, which is opposes our direction of motion here to the right.
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So if you want to know it when it's just starting to move, we want to find it when it's balanced so that as soon as it increases, we are moving the box.
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That's our x direction, so our forces in our y direction also have a net value of zero because it's not accelerating in the y direction at all.
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So back to our diagram, we have our normal in the positive direction, our gravity pointing downward, and we have a portion of our applied force in the y direction given by sign...