Relativity of Orientation & Origin

  • Thread starter knowLittle
  • Start date
  • #1
307
0

Homework Statement


At time t=0, a block is released from point O on the slope shown in the figure.
The block accelerates down the slope, overcoming sliding friction.

a.) Choose axes 0xy as shown, and solve the equation ##\Sigma F = m a## into its x and y components.
Hence find the block's position (x,y) as a function of time, and the time it takes to reach the bottom.

b.) Carry out the solution using the axes Ox'y', with Ox' horizontal and Oy' vertical, and show that you get the same final answer. Explain why the solution using these axes is less convenient.


Homework Equations


distance =v*t


The Attempt at a Solution


Normal force = mg

## m*g* cos(\theta) = ##force perpendicular to the base not the ramp

y=0 in the 0xy frame.
I know that to obtain x I need to subtract the friction force upwards.

However, I am having problems finding x.
So, Fnet_x = Fx - F_friction

Fx= hypotenuse*cos##(\theta)##

Help.
 

Attachments

  • f3.jpg
    f3.jpg
    3.8 KB · Views: 378

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
965

Homework Statement


At time t=0, a block is released from point O on the slope shown in the figure.
The block accelerates down the slope, overcoming sliding friction.

a.) Choose axes 0xy as shown, and solve the equation ##\Sigma F = m a## into its x and y components.
Hence find the block's position (x,y) as a function of time, and the time it takes to reach the bottom.

b.) Carry out the solution using the axes Ox'y', with Ox' horizontal and Oy' vertical, and show that you get the same final answer. Explain why the solution using these axes is less convenient.


Homework Equations


distance =v*t


The Attempt at a Solution


Normal force = mg

## m*g* cos(\theta) = ##force perpendicular to the base not the ramp
You have these backwards. mg is the force of gravity- straight downwards. It is perpendicular to the base, not the ramp. [itex]mg cos(\theta)[/itex] is perpendicular to the ramp, not the base.

y=0 in the 0xy frame.
I know that to obtain x I need to subtract the friction force upwards.

However, I am having problems finding x.
So, Fnet_x = Fx - F_friction

Fx= hypotenuse*cos##(\theta)##

Help.
 
  • #3
307
0
You have these backwards. mg is the force of gravity- straight downwards. It is perpendicular to the base, not the ramp. [itex]mg cos(\theta)[/itex] is perpendicular to the ramp, not the base.

Ok, thank you. I remember things better now.

## F_{net} = m*a ##
## F_x = m*g*sin(\theta)##
## F_{\mu} = \mu * F_N##

##F_{net}= F_x- F_{\mu}=m*a##
## a = g(sin(\theta) - \mu(cos(\theta)))##
Finally,
## S(t)= (1/2)g(sin(\theta) - \mu(cos(\theta)))* t^2 ##
 
Last edited:

Related Threads on Relativity of Orientation & Origin

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
20
Views
3K
Replies
27
Views
3K
  • Last Post
Replies
5
Views
6K
Replies
1
Views
2K
Replies
3
Views
505
  • Last Post
Replies
2
Views
3K
Replies
1
Views
1K
Top