• Support PF! Buy your school textbooks, materials and every day products Here!

Block pushed up wall

  • Thread starter splinter
  • Start date
  • #1
10
0
Hey everyone, my first post here, hope it's ok that it's a question...
I've got this physics problem that is absolutely stumping me, is abnormal from all the other forces problems I've done before, and uk is throwing me for a loop too...here it is:

6. [SFHS99 5.P.60.] A 5.5 kg block is pushed 3.1 m at a constant speed up a vertical wall by a constant force applied at an angle of 30.0° with the horizontal, as shown in Figure 5-23.
Figure 5-23
Assume that the coefficient of kinetic friction between the block and the wall is 0.30.

(a) Determine the work done by the force on the block
___________________ J
(b) Determine the work done by gravity on the block.
___________________ J
(c) Determine the normal force between the block and the wall.
___________________ N

Any help or hints you guys can give me would be greatly appreciated...this one has me stumped.
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
1
You know there will be a normal force, find out which will it be equal to, remember the block isn't moving the x-axis so the Fx sum is zero.
Also Remember Newton's 1st Law

[tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{v} = constant[/tex]

Think in components of the applied force.
 
Last edited:
  • #3
10
0
Ok so I tried taking the gravitational force, 54.96N, and finding the horizontal component of that. I drew it as a triangle with 54.96N as the y leg, then did 53.96/tan 30 to find the x leg of the triangle. This gave me 93.46N, which my online homework is saying is wrong! Still confused =\
 
  • #4
Pyrrhus
Homework Helper
2,178
1
Applying Newton's 1st Law

[tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{v} = constant[/tex]

on x-axis

[tex] Fcos\theta - N = 0 [/tex]

[tex] Fcos\theta = N [/tex]

On y-axis

[tex] Fsin\theta - mg - F_{f} = 0 [/tex]

[tex] F_{f} = \mu N [/tex]

[tex] Fsin\theta - mg -\mu Fcos\theta = 0 [/tex]

Gravity will have a negative work because it forms 180 degrees with the displacement vector, and [itex] Fsin\theta [/itex] will do the work pushing the block up.

Also remember Work definition

[tex] W = \vec{F} \cdot \vec{r} [/tex]

where W is equal to

[tex] W = |\vec{F}||\vec{r}|cos\theta [/tex]

where [itex] \theta [/itex] is the angle between the radius vector and the force.
 
Last edited:
  • #5
10
0
ok I've gotten the Fapp and the Fn, now I'm just having trouble with the Ff. I used the equation Ff = uN, plugging in my normal force of 194.5 and u of .3, then multiplied that by 3.1 to get work, and entered the negative of that, but it's still telling me its incorrect. Don't understand what I'm doing wrong!
 
  • #6
10
0
Didn't see your new post and edit, might be able to get it with that.
 
  • #7
Pyrrhus
Homework Helper
2,178
1
(a) Determine the work done by the force on the block
___________________ J
(b) Determine the work done by gravity on the block.
___________________ J
(c) Determine the normal force between the block and the wall.
___________________ N

Read the questions again, it does not ask for work done by friction... I also decided to delete my energetic solution, because it wasn't needed.
 
  • #8
10
0
Whoops it was work done by gravity...easy just mgd. doh! Thanks for the help i got them all now.
 
  • #9
Pyrrhus
Homework Helper
2,178
1
It was a pleasure to be of help, and welcome to PF!
 

Related Threads on Block pushed up wall

  • Last Post
2
Replies
39
Views
13K
  • Last Post
Replies
10
Views
18K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
760
Replies
4
Views
17K
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
17
Views
3K
  • Last Post
Replies
2
Views
3K
Replies
2
Views
21K
  • Last Post
Replies
2
Views
4K
Top