Horizontal component of force without angle given?

In summary, the person is applying a constant force of 453N to the 60kg block, which is causing it to slide along the ground with a coefficient of friction of μk=0.60.
  • #1
Hanyuu
3
0

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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  • #2
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:
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
Fnet=ma, since acceleration is 0, then Fnet is also zero. Friction force also affects horizontal force.

Solved the problem, thanks!
 

1. What is the formula for calculating the horizontal component of force without angle given?

The formula for calculating the horizontal component of force without angle given is Fx = F * cos θ, where F is the magnitude of the force and θ is the angle between the force vector and the horizontal axis.

2. Can the horizontal component of force be negative?

Yes, the horizontal component of force can be negative. This means that the force is acting in the opposite direction of the positive horizontal axis.

3. How is the horizontal component of force related to the vertical component of force?

The horizontal component of force and the vertical component of force are both components of the overall force vector. They are related through the Pythagorean theorem, where the magnitude of the overall force is equal to the square root of the sum of the squares of the horizontal and vertical components.

4. What is the unit of measurement for the horizontal component of force?

The unit of measurement for the horizontal component of force is the same as the overall force, which is typically measured in newtons (N) in the metric system or pounds (lbs) in the imperial system.

5. How does the angle of the force affect the horizontal component of force?

The angle of the force affects the horizontal component of force through the cosine function. As the angle increases, the horizontal component decreases and vice versa. When the force is acting at a 90 degree angle to the horizontal axis, the horizontal component will be zero.

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