Horizontal component of force without angle given?

AI Thread Summary
A person is dragging a 60kg block at a constant velocity on flat ground, with an upward pull force component of 453N and a coefficient of friction of 0.60. The discussion emphasizes that the problem can be solved without an angle by recognizing that the net force on the block is zero due to constant velocity. Participants suggest using a free body diagram to analyze the forces acting on the block, including gravitational and frictional forces. The importance of understanding that the block is not accelerating is highlighted, leading to the conclusion that the horizontal component of the pull force can be calculated. The problem was ultimately solved by applying these principles.
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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