Horizontal component of force without angle given?

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Homework Help Overview

The discussion revolves around a physics problem involving a 60kg block being dragged at a constant velocity over flat ground. The problem requires calculating the horizontal component of the pulling force, given the upward component of the force and the coefficient of friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express confusion about the necessity of an angle to determine the horizontal component of the force. There are discussions about the implications of the block moving at constant velocity and the role of free body diagrams in visualizing forces.

Discussion Status

Participants have engaged in clarifying the conditions of the problem, particularly the significance of the block's constant velocity, which implies that the net horizontal force is zero. Some guidance has been offered regarding the use of free body diagrams to analyze the forces acting on the block.

Contextual Notes

There is an emphasis on the lack of an angle in the problem statement, which some participants note as a potential barrier to finding the solution. The discussion also highlights the importance of understanding the relationship between the forces involved, particularly friction and the applied force.

Hanyuu
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Homework Statement


A person is dragging a 60kg block at a constant velocity over flat ground. The force applied by the person to the block has components that are up and to the right. The coefficient of friction for sliding the block over the ground is μk=0.60.

Calculate the rightward component of the person's pull force assuming the upward component of the pull force is 453N


Homework Equations


F vertical = Fsinθ
F horizontal = Fcosθ


The Attempt at a Solution


I'm confused about how to approach this. Shouldn't there be an angle of force given in order to solve this?
 
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Hello Hanyuu,

Welcome to Physics Forums!
Hanyuu said:

Homework Statement


A person is dragging a 60kg block at a constant velocity over flat ground. The force applied by the person to the block has components that are up and to the right. The coefficient of friction for sliding the block over the ground is μk=0.60.

Calculate the rightward component of the person's pull force assuming the upward component of the pull force is 453N

Homework Equations


F vertical = Fsinθ
F horizontal = Fcosθ

The Attempt at a Solution


I'm confused about how to approach this. Shouldn't there be an angle of force given in
You can solve this without being given an angle. Drawing a free body diagram would be a good way to start.

There is one critical piece of information that you don't want to neglect: The block is being dragged with a constant velocity. In other words, the block is not accelerating. (And it's being dragged over flat ground too.)
 
Last edited:
Hanyuu said:

Homework Statement


A person is dragging a 60kg block at a constant velocity over flat ground. The force applied by the person to the block has components that are up and to the right. The coefficient of friction for sliding the block over the ground is μk=0.60.

Calculate the rightward component of the person's pull force assuming the upward component of the pull force is 453N


Homework Equations


F vertical = Fsinθ
F horizontal = Fcosθ


The Attempt at a Solution


I'm confused about how to approach this. Shouldn't there be an angle of force given in order to solve this?

What are your thoughts about the net horizontal force on the block?
 
gneill said:
What are your thoughts about the net horizontal force on the block?

Ah, Fnet should be 0, right?

collinsmark said:
Hello Hanyuu,

Welcome to Physics Forums!

You can solve this without being given an angle. Drawing a free body diagram would be a good way to start.

There is on critical piece of information that you don't want to neglect: The block is being dragged with a constant velocity. In other words, the block is not accelerating. (And it's being dragged over flat ground too.)

So Fnet would be adding up Fg, Ff, Fn and the two components?
 
Hanyuu said:
Ah, Fnet should be 0, right?
Yes; What is your argument for that?

If you draw a Free Body Diagram for the block, what comprises the contributions to the net horizontal force? Which do you have enough information to compute?
 
Fnet=ma, since acceleration is 0, then Fnet is also zero. Friction force also affects horizontal force.

Solved the problem, thanks!
 

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