Motion and Forces for a block held against a wall

AI Thread Summary
The discussion revolves around a physics problem involving a block held against a wall by a horizontal force, where the user is struggling with calculating the correct frictional forces and accelerations for various applied forces. The user initially provided incorrect answers for specific scenarios, particularly for cases (c) and (i), and sought clarification on the signs for frictional forces. It was established that the direction of friction is negative when opposing the motion, and the user learned that the correct sign convention is crucial for accurate calculations. The correct frictional force for scenario (c) was determined to be -26 N, indicating downward friction. Overall, the conversation emphasizes the importance of understanding force directions and sign conventions in solving physics problems.
maria1998
Messages
3
Reaction score
0

Homework Statement



Hello, I have attached a physics problem that I solved yet some answers (in red+the first answer) are wrong and I can't find out why...

Homework Equations

upload_2016-9-14_11-48-46.png
[/B]

The Attempt at a Solution


upload_2016-9-14_11-49-24.png

[/B]
 
Physics news on Phys.org
Hello Maria, :welcome:

Your question is clear enough, but PF works a little different (see guidelines). The picture can serve as a problem statement, but you still need to post the relevant equations and your attempt at solution (not just the outcome). So, show what you did to calculate the numbers and we'll help you further ...
 
the problem is: In the figure, a block weighing 23.0 N is held at rest against a vertical wall by a horizontal force https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2528958entrance1_N10036.mml?size=14&ver=1473845185939 of magnitude 61 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second force https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2528958entrance1_N1005C.mml?size=14&ver=1473845185939 is applied to the block and directed parallel to the wall with these magnitudes and directions: (a)35 N, up, (b)13 N, up, (c)49 N, up, (d)61 N, up, (e)9.0 N, down, and (f)18 N, down. In each experiment, what is the frictional force on the block, including sign? Take the direction up the wall as positive, and down the wall as negative. Next, calculate the acceleration, including sign, of the block in each case. Note that acceleration is zero if the block does not move.
(g) What is the acceleration in (a)?
(h) What is the acceleration in (b)?
(i) What is the acceleration in (c)?
(j) What is the acceleration in (d)?
(k) What is the acceleration in (e)?
(l) What is the acceleration in (f)?

I got a wrong answer for c and i here's what i did:
for the c : sum of forces=0 which means P-W-f=0 ; 49-23=26 so f=26 but it is wrong
and for the i: a=sum of forces/m and m=W/9.8 I got -27.35 but it is wrong
 
Last edited by a moderator:
I didn't intend for you to retype the whole problem statement, sorry.

The correct sign for up and down follows from b) which is counted as correct answer: + 10 N means 10 N up. Hence the cross at answer a, I assume.

What is the direction of the friction in (c) ? Which way is your f positive ?

I expect the cross at c Units to disappear once the numerical value is right ?
 
it's okay i just copy pasted it
anyway i found the right answers but i still don t understand when f is positive or negative...
 
In (b) 23 N is down, 13 N is up, so the sum is zero with 10 N up. +10 is counted correct, so up = positive in this exercise.

For (c) you would get 49 N - 23 N (down is negative) + f = 0 leaves -26 N for f. Apparently downwards. |-26| N is still below the max friction force, so -26 N should be the right answer.
 

Similar threads

Back
Top