Calculating the Force Needed to Pull a Block

AI Thread Summary
A block weighing 455 N is being pulled at a constant velocity with a force at a 30-degree angle above the horizontal, while experiencing a friction force of 1163 N. The discussion highlights the need to resolve forces correctly, emphasizing that the weight of the block, 445 N, acts vertically downward and should not be resolved into the horizontal direction. Participants suggest drawing a free body diagram to visualize the forces acting on the block, which includes the pulling force, friction, and gravity. The goal is to calculate the magnitude of the pulling force required to maintain constant velocity despite friction. The conversation reflects a collaborative effort to clarify the problem-solving approach.
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Homework Statement


A block with eight of 455 N is being pulled at constant velocity by a force directed at 30 degrees above the horizontal. The friction force on the crate is 1163 N. What is the magnitude of the pulling force?

Homework Equations


Ft= kx
Fnet= ma


The Attempt at a Solution


I tried to use the component method to find the force in the x direction and then so that I could find the pulling force but it doesn't work!
 
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What, specifically, did you try?
 
I tried using 445 N cos 30 degree to find the vector on the x direction, that's how I was taught!
 
445 N is the weight of the block, which acts vertically downwards, so I'm not sure why you're resolving that into the x direction.

You have several pieces of information: the block has a weight of 445N; it is traveling at constant velocity; the pulling force is 30 degrees above the horizontal; the friction acting on the box is 1163N.

So, first things first, I would draw a free body diagram including all the forces. Can you do this?
 
sure, I do

and I did draw it

so to the left is the friction force

and 30 degrees above the horizontal is the acting force

the gravity is 9.81 and from there I can find the mass of the object

and I have to find the pulling force

thank you very much for you help

I will try my best and see what I get
 
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