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Problem 98 Hard Difficulty

mmh Two forces, $\overrightarrow{\mathbf{F}}_{1}$ and $\overrightarrow{\mathbf{F}}_{2},$ act on the $7.00-\mathrm{kg}$ block shown in the drawing. The magnitudes of the forces are $F_{1}=59.0 \mathrm{N}$ and $F_{2}=33.0 \mathrm{N}$ . What is the horizontal acceleration (magnitude and direction) of the block?

Answer

1.83 $\mathrm{m} / \mathrm{s}^{2}$ to the left

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Video Transcript

to solve this question and have to use a nuclear second law. And for that you choose the following reference frame vertical access that I will call why and a horizontal axis that I'll call X begin by noticing that the force F one isn't pointing to the axe direction or to the Y direction, but to both directions. So we have to decompose F one in its components. There is one horizontal component. We've F one component X and the vertical component, which is F one component white. Then we can see the following. This angle is an angle off 70 degrees, then these angle is an angle off 90 degrees. Reform. The only possibility for the intern angle these one is that these is 20 degrees. Moreover, this is also 90 degrees and then we again have an angle off 70 degrees appear. Using this angles, we can calculate both components off F one. It was as follows by using the sign off 20 degrees, which is given by in this context the opposite side of triangle. So our foreign component X, divided by F one. We are calculating as one component x because then F one component exit physicals to f times this sign off 20 degrees, then, using the co sign off 20 degrees, we can calculate F one component. Why? Because the coastline involves the address inside off the triangle, which is F one way, then the to sign off 20 is given by F one component. Why divided by f Shoot F one component white music was to f times the co sign off 20 degrees. There is a detail here, actually. Um, minor mistake. That is the following. I'm working with F one, but I have been writing f So let me correct this by adding subscript one. So now we knew Booth F one on its components, and now we're able to apply Newton's second law. Applying it to the X axis tells us that the Net force next direction is it caused the mass off the box times. It's horizontal acceleration, which is acceleration in the X direction, the next forcing next Directions. Composed by F two and F one component Web X. But F one component acts is pointing to the right so f one x minus. F two Is it close to M times a component X But we know that F one X is given by F one times the sign off 20 degrees. Then there's his minus. After on these easy goes to AM times a X. Therefore, the acceleration next direction is given by F one times the sign off 20 degrees minus F troop divided by the mass off the box, plugging in the values that were given by the problem. We got 59 times the sign off 20 minus 33 and this is divided by seven, resulting in the horizontal acceleration off approximately minus 1.83 meters per second squared Notice the minus sign. It tells us that the acceleration is pointing to the negative direction off our exactness or you know the words. It's 1.83 meters per second squared to the left, so it's meeting to the left.

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