Force Vectors: In a warehouse a box is being pulled up a ramp that has a 15 degree incline....

[Randy Hammond]

I have come across a problem that is giving me troubles and need explanations behind the math concepts. I am a HS math teacher that has not done vectors in a very long time and need a little refresher. Most of the problems seem straight forward, like 46, but I am stuck on 47. Since problem 47 refers to number 46, I have included them both in the following attachment. Below is an overview of the problem.

In a warehouse a box is being pulled up a ramp that has a 15 degree incline. The rope, vector w, that is pulling this box makes a 33 degree angle from the box with the horizontal. Find the force w needed in order for the component of the force parallel to the inclined plane to be 2.5 lb.

The answer is supposed to be in component form and is suggested to be <2.20, 1.43>.
Thank you for the help in advanced.

First Attachment:

 

[mmm4444bot]

Second Attachment:

 

[The Highlander]

I have come across a problem that is giving me troubles and need explanations behind the math concepts. I am a HS math teacher that has not done vectors in a very long time and need a little refresher. Most of the problems seem straight forward, like 46, but I am stuck on 47. Since problem 47 refers to number 46, I have included them both in the following attachment. Below is an overview of the problem.

I suspect that (in Q.46) you just found 2.5cos15° & 2.5sin15° and presumed, therefore, that the question was quite straightforward but what did you put for part (b)? The Physics might not be quite as "straight forward" (sic) as you think.

I haven't tried to decipher what you have attempted in your second attachment but a cursory glance tells me that your approach is wrong (and the Physics worse, lol). However, if we ignore the Physics for a moment, the Maths is, again, quite simple (the Physics not so) and my diagram illustrates how to solve the problem in a way you should be very comfortable with.

The 2.5 lb force is a component of W but what is W?
Clearly (from the diagram), 2.5 = Wcos18° ⇒ W ≈ 2.63 lb \(\displaystyle \left(\frac{2.5}{cos18°}\right)\)
So, now you know W, you can easily split it into its (approximate) horizontal & vertical components as: 2.63cos33° & 2.63sin33°.

NB: Don't round the value for W (to 2.63) until you have calculated the component values (or you will get 2.21 instead of 2.20).

Hope that helps.

 

[Randy Hammond]

I suspect that (in Q.46) you just found 2.5cos15° & 2.5sin15° and presumed, therefore, that the question was quite straightforward but what did you put for part (b)? The Physics might not be quite as "straight forward" (sic) as you think.

I haven't tried to decipher what you have attempted in your second attachment but a cursory glance tells me that your approach is wrong (and the Physics worse, lol). However, if we ignore the Physics for a moment, the Maths is, again, quite simple (the Physics not so) and my diagram illustrates how to solve the problem in a way you should be very comfortable with.

View attachment 37926The 2.5 lb force is a component of W but what is W?
Clearly (from the diagram), 2.5 = Wcos18° ⇒ W ≈ 2.63 lb \(\displaystyle \left(\frac{2.5}{cos18°}\right)\)
So, now you know W, you can easily split it into its (approximate) horizontal & vertical components as: 2.63cos33° & 2.63sin33°.

NB: Don't round the value for W (to 2.63) until you have calculated the component values (or you will get 2.21 instead of 2.20).

Hope that helps.

Thank you! This does help and clarifies much more. Thank you again.

 

[The Highlander]

Thank you! This does help and clarifies much more. Thank you again.

You're very welcome.