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Find the resultant of 2 tensile forces

Find the resultant of two tensile forces 2p and 3p acting on a particles if the angle between them is 60 degrees.

How do I go about answering this?

Reply 1

Smack

Volunteer Team

Original post by lilminster

Find the resultant of two tensile forces 2p and 3p acting on a particles if the angle between them is 60 degrees.

How do I go about answering this?

What does 3p and 3p mean? Is there a diagram?

Reply 2

lilminster

No diagram, the answer is 4.36p at 23.4 degrees to the 3p force. I know I have to use the following: a^2 = b^2 + c^2 + 2bc cos A but not sure why it's + 2bc rather than - 2bc

Reply 3

titfortat

Original post by lilminster

No diagram, the answer is 4.36p at 23.4 degrees to the 3p force. I know I have to use the following: a^2 = b^2 + c^2 + 2bc cos A but not sure why it's + 2bc rather than - 2bc

That's odd. It should be -2bc, does it say on the mark scheme to use +2bc?

Reply 4

lilminster

Nope, it just says the answer on the mark scheme :/ I've found a similar question using that same +2bc formula:
http://www.askiitians.com/forums/Mechanics/10/56415/vectors.htm

Reply 5

pleasedtobeatyou

Original post by lilminster

Find the resultant of two tensile forces 2p and 3p acting on a particles if the angle between them is 60 degrees.

How do I go about answering this?

I have run through the calculations with the traditional interpretation of the cosine rule i.e. with

-2bc

and have ended up with the mark scheme answers.

R^2 = (2P)^2 + (3P)^2 - 2(3P)(2P)cos(120^{\circ})[br]R^2 = 19P^2[br]\implies R = 4.36P

.

To find the angle

\theta

between the resultant and the 3P force, use the fact that the horizontal component of the resultant force and it's two sub-component forces must be the same.

Rcos(\theta) = 3P + 2Pcos(60^{\circ})[br]\implies \theta = 23.4^{\circ}

Reply 6

lilminster

Original post by pleasedtobeatyou

I have run through the calculations with the traditional interpretation of the cosine rule i.e. with

-2bc

and have ended up with the mark scheme answers.

R^2 = (2P)^2 + (3P)^2 - 2(3P)(2P)cos(120^{\circ})[br]R^2 = 19P^2[br]\implies R = 4.36P

.

To find the angle

\theta

between the resultant and the 3P force, use the fact that the horizontal component of the resultant force and it's two sub-component forces must be the same.

Rcos(\theta) = 3P + 2Pcos(60^{\circ})[br]\implies \theta = 23.4^{\circ}

Thank you so much!

Reply 7

Smack

Volunteer Team

Original post by lilminster

No diagram, the answer is 4.36p at 23.4 degrees to the 3p force. I know I have to use the following: a^2 = b^2 + c^2 + 2bc cos A but not sure why it's + 2bc rather than - 2bc

Draw a diagram, and resolve the forces based on said diagram. Just copying a formula isn't a good idea, especially if there may be a said typo or error in the formula. Note how @pleasedtobeatyou does it using the cosine rule and arrives at the correct solution.