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Problem 119 Hard Difficulty

The drawing shows a $25.0-\mathrm{kg}$ crate that is initially at rest. Note that the view is one looking down on the top of the crate. Two forces, $\overrightarrow{\mathbf{F}}_{1}$ and $\overrightarrow{\mathbf{F}}_{2},$ are applied to the crate, and it begins to move. The coefficient of kinetic friction between the crate and the floor is $\mu_{\mathrm{k}}=0.350 .$ Determine the magnitude and direction (relative to the $x$ axis) of the acceleration of the crate.

Answer

1.61 $\mathrm{m} / \mathrm{s}^{2}$
$34.6^{\circ}$ abov

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Chew C.

October 12, 2020

A person pushes on a 72.6 kg refrigerator with a horizontal force of -267 N

CA

Catherine A.

October 27, 2020

Dading C., thanks this was super helpful.

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Catherine A.

October 27, 2020

This will help alot with my midterm

Video Transcript

So the drawing, which looks something like this, You have F two Is 54. Newton's pointing to the right. F one is 80 Newtons pointed up into the left, this is the y axis and that is the X axis. Yeah, Okay, so we need to determine the magnitude in the direction of the acceleration of the great, So notice that if you're pulling to the right and up into the right, the direction of motion is going to be to the left. So our acceleration needs to be positive to find this net force. It's the square root of F one x plus F two, X squared plus F one Y squared. There's no F T Y term, because FT was only in the X direction. So 88 F one X is 88 co sign of 55 degrees from the figure. So, plugging this in and we get a force of 126 nine newtons. So, looking at the son of all of the forces, you have friction pulling you in the opposite direction. So the sum of all of the forces is mass times acceleration From Newton's 2nd law. So this is the forest minus the friction, which is um UK times the normal force here. The normal force is equal to the weight. So um U k mg. We know the U K. We know and jesus constant. We just found F. So actually the masses cancelled, so we don't need to worry about the masses. And we get that the acceleration Is equal to 1.65 meters per second squared and the angle is equal to the inverse tangent of the Y component of the forces, which is F one way over the X component, which is F one X plus F two. So we get an angle of 34.6°