Find the Force Needed: Wood Box Sliding Down a Wall

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Homework Help Overview

The problem involves a 2.7 kg wood box sliding down a vertical wall while being pushed at a 35° angle. The objective is to determine the magnitude of the force required to maintain a constant speed as the box descends.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of net force in relation to constant speed, questioning the balance of forces acting on the box. There are attempts to resolve the applied force into vertical and horizontal components, and some participants express confusion about the necessary formulas.

Discussion Status

Multiple interpretations of the problem are being explored, with participants offering insights into the relationship between gravitational force and the applied force. Some guidance has been provided regarding the need for net force to equal zero for constant velocity, but no consensus on the correct approach or final answer has been reached.

Contextual Notes

Friction is not mentioned in the problem statement, leading to questions about its relevance. The angle of application for the force is also discussed, with some participants clarifying the orientation of the angle in relation to the vertical.

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Homework Statement


A m = 2.7 kg wood box slides down a vertical wood wall while you push on it at a θ = 35° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?


Homework Equations





The Attempt at a Solution



I got 64.5766 N for the answer. Don't know how to do it.
 

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Do you know what the net force on the box is equal to if it's traveling at a constant speed?

What force is the box moving downwards with? (if your hands aren't there).

just some things to consider

remember that for it to not be accelerating up or down the force you push it up with (vertically) must be equal to the force it's coming down with. think about splitting the 35 degree force into vertical and horizontal forcess
 
don't really understand that... could you tell me what formulas to use? thanks
 
okay, do you understand that in order for something to be static, or traveling with a constant velocity, there must be no net force acting on it?

i think it's far more important to understand what to do here than simply use formulae.

also do you know how to resolve forces into vertical and horizontal?
 
yeah I understand the no net force thing...and resolving vertical and horizontal forces...using sin and cos. yes...
 
Okay, so you must therefore realize that if it's moving at a constant speed (downwards) it's net force will be 0.

therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

your equation will look something like so;

mg - Fcos(theta) = 0

is friction mentioned at all?
 
Last edited:
I got 29.2799 for the final answer. it is wrong
 
that's not terribley helpful...

what working out did you use to get that? if friction mentioned at all?
 
I have this exact same question. Friction is not mentioned at all. He stated the exact wording already. I need help on this question as well. Is it ok to hijack this thread, or do I have to start a new one?
 
  • #10
No that's fine, it's the same question.

Hmm when i tried it I got 32.3N, is that the correct answer? If it is, see my post above, which says this

Okay, so you must therefore realize that if it's moving at a constant speed (downwards) it's net force will be 0.

therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

your equation will look something like so;

mg - Fsin(theta) = 0if not tell me, the answer may be wrong
 
Last edited:
  • #11
Actually its the sine component that's balancing the weight of the body. So your answer might be around 47 or so, I reckon.
 
  • #12
sArGe99 said:
Actually its the sine component that's balancing the weight of the body. So your answer might be around 47 or so, I reckon.

Hm, would presume they mean 35 degrees from the vertical, not horizontal.

edit: oh right just saw the picture attached, my bad :P
 

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