Solving Forces and Acceleration Problems in Physics - Study Group Help

  • Thread starter Thread starter caity232
  • Start date Start date
  • Tags Tags
    Force Newton
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving forces and acceleration. The original poster presents a scenario where two forces are applied to a car, and they seek to determine the resultant force and the acceleration of the car based on its mass.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to break down the forces into their components and sum these components to find the resultant force. There are questions about the calculations of the components and the application of the Pythagorean theorem to find the resultant's magnitude.

Discussion Status

Some participants are providing guidance on how to approach the problem by suggesting the use of a coordinate system and the importance of treating acceleration as a vector. There is an ongoing exploration of the calculations, with some participants expressing confusion over the results and seeking further clarification.

Contextual Notes

Participants are working under the constraints of a homework assignment, and there is a mention of answers provided in the textbook, which are not aligning with their calculations. The discussion reflects uncertainty about the correct application of vector addition and component analysis.

caity232
Hello, I am a junior in High School and some of my friends in my physics class and I have a study group for our weekly bookwork that our teacher assigns. We have tried all week to figure out this one problem but to no avail. The problem is as follows:
Two forces are applied to a car in an effort to move it. A)What is the resultant of these two forces? B) If
the car has a mass of 3000 Kg, what acceleration does
it have?

the diagram of the car shows that it is being pulled
in two different directions one arrow is pointing
northeast with a force of 400 Newtons at an angle of
30 degrees, and the other direction is northwest at an angle
of 10 degrees and a force of 450 Newtons.

I tried using 400cos30 and 450cos10, but it doesn't work. The answers are in the back of the book, but I'm looking for how to do the math and I don't want to just copy the answers down because I would like to know how to solve this. If it helps, the answers are:
A) 789 Newtons at 8.8 degrees to the right of forward direction
B) 0.266 m/s^2 in direcrion of resultant force
Thank you in advance, and please help me with the math in this problem.
 
Physics news on Phys.org
I tried using 400cos30 and 450cos10, but it doesn't work.
Close, very close.

You have to remember that acceleration is also a vector, and so also need to be seen in terms of components.

So, what you want to do is first to define a coordinate system. Put North as y, say, and East as x.

Then, you 400cos30, 400sin30 and so on to put to two forces in component form along the two directions, and then sum the x components and sum the y components. Now you have the resultant force in component form.

From the presentation of the answer, you need then to calculate the angle and magnitude of this force. For that use pythagoras and arctan.

For acceleration, it's an easy matter of using f=ma on the magnitude.
 
when you add the x and y components you get 1067.71 ... when i did 400cos30 and 400sin30 i got 546.41 and when i did 450cos10 and 450sin10 i got 1067.71... The answer in the back of the book in 789? Can I have some further help with this?
 
Why are you adding the x and y components? The length of a vector is not the length of its components. Once you have found the components and add "x to x" and "y to y" to find the resultant, you can use "Pythagorean Theorem" to find the length.
 
it still doesn't give you 789
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K
Replies
43
Views
4K
Replies
20
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K