Video: Determining the Component Form of Net Acceleration of an Object Acted on by Multiple Forces

The body shown being acted on by three forces has a mass of 10.0 kg. Find the acceleration of the body.

07:11

Video Transcript

The body shown being acted on by three forces has a mass of 10.0 kilograms. Find the acceleration of the body.

Looking at the diagram given, we see that the mass ๐‘š is being acted on by ๐น one, ๐น two, and ๐น three. To solve for the acceleration of mass ๐‘š, weโ€™ll want to know what are the ๐‘ฅ- and ๐‘ฆ-components of these three forces acting on the mass. We can do that and organize our information more easily using a table. In this table, weโ€™ll have our three forces: ๐น one, ๐น two, and ๐น three as well as the components of those forces: ๐น sub ๐‘ฅ and ๐น sub ๐‘ฆ. And what weโ€™re trying to do is after figuring out the ๐‘ฅ- and ๐‘ฆ-components of each of the three forces, we want to add them together to find the total ๐‘ฅ- and the total ๐‘ฆ-component of the force acting on the mass ๐‘š.

So here is our table that weโ€™ll fill in and again we want to look at the ๐‘ฅ- and ๐‘ฆ-components of the three forces acting on ๐‘š. So letโ€™s start with ๐น one. As we look at ๐น one in the diagram, we see that it has no horizontal component; itโ€™s all vertical. So the ๐‘ฅ-component of ๐น one is zero. Moving on to its ๐‘ฆ-component, we see ๐น one points in the negative ๐‘ฆ-direction so that ๐‘ฆ-component of the force is negative 10.0 newtons.

Filling that into the table, we move on to ๐น two. Now, ๐น two, unlike ๐น one, has both ๐‘ฅ- and ๐‘ฆ-components. And to figure those out, we want to use the fact that we can draw a horizontal dotted line between the dotted vertical line we see and the end of ๐น two. So letโ€™s draw that in now. What we see with that line drawn in is that our two dotted lines are at right angles to one another and that these dotted lines together with ๐น two form a right triangle.

So we can solve for the ๐‘ฅ- and ๐‘ฆ-components of ๐น two using that fact. The ๐‘ฅ-component of ๐น two is equal to the magnitude of ๐น two, 20.0 newtons, multiplied by the sine of 37 degrees. You maybe more used to seeing the cosine associated with horizontal or ๐‘ฅ-direction motion or forces, but in this case because of the particular way weโ€™ve drawn our triangle, indeed weโ€™re working with the sine function to find that horizontal component of ๐น two.

With that ๐‘ฅ-component of ๐น two added to our table, now letโ€™s work on the ๐‘ฆ-component. That will be equal to again the magnitude of ๐น two, 20.0 newtons, multiplied this time by the cosine of 37 degrees because the side of our triangle we want to solve for is the side adjacent to our 37-degree angle. Letโ€™s fill that value in in our table.

Now, we move on to ๐น three, the last force of the three acting on mass ๐‘š. We see that just like with ๐น two, we can draw in a dotted horizontal line that joins the end or the arrowhead of ๐น three with the dotted vertical line weโ€™ve drawn in. That line shown here forms the opposite side of our triangle.

So the ๐‘ฅ-component of ๐น three is equal to the magnitude of ๐น three multiplied again by the sine of 37 degrees. But notice that in this case that value is negative because the ๐‘ฅ-component of ๐น three points in the negative ๐‘ฅ-direction. So overall that will be negative 10.0 newtons multiplied by the sine of 37 degrees. Then finally, the ๐‘ฆ-component of ๐น three is equal to positive 10.0 newtons multiplied by the cosine of 37 degrees.

Great! Now our table is all filled out and weโ€™re ready to add up the ๐‘ฅ- and ๐‘ฆ-columns to solve for the ๐‘ฅ- and ๐‘ฆ-components of the net force โ€” the total force โ€” acting on mass ๐‘š. Adding up the three values in our ๐‘ฅ-column, we get an answer of 6.02 newtons. And in the ๐‘ฆ-direction, negative 10 newtons plus 20 newtons times the cosine of 37 degrees plus 10 newtons times the cosine of 37 degrees equals 14.0 newtons.

Now again, these are the components of the net force that acts on ๐‘š. So letโ€™s write it out and make it official; weโ€™ll call that net force ๐น sub net. And we now know that that force is equal to 6.02 ๐‘– plus 14.0 ๐‘— newtons. And remember that this is all in service to figuring out what is the acceleration that the mass ๐‘š takes on under the influence of all three of these forces.

So weโ€™ve solved for that net force, resulting from ๐น one, ๐น two, and ๐น three. And weโ€™ve been given the mass of our object, 10.0 kilograms, and wanna solve for its acceleration. One of Newtonโ€™s laws of motion, the second law connects all three of these values. The second law says that the net force on a mass ๐‘š is equal to that mass multiplied by its acceleration. We can use this law to solve for acceleration. Letโ€™s start by dividing our net force by the mass of our object ๐‘š.

Now when we do this, what weโ€™ve done has created a fraction that is equal to the acceleration of our mass. Thatโ€™s true by Newtonโ€™s second law. So now we can plug in for our mass ๐‘š, the mass of 10.0 kilograms. Having done that, we now divide these numbers using our calculator to solve for the net acceleration of the mass ๐‘š.

And we find that that net acceleration that mass ๐‘š experiences under the influence of these three forces is 0.602 ๐‘– plus 1.40 ๐‘— meters per second squared. That is the net acceleration of mass ๐‘š.

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