Box dragged along floor at an angle, calculating normal and frictional force

In summary, the box is dragged along a rough but level floor by a rope at a 30o angle to the horizontal and is speeding up. The frictional force of the floor is 513 Newtons, and the horizontal component of the 150 N force is 129.9 N.
  • #1
revere21
10
0

Homework Statement


A 60 kg box is dragged along a rough but level floor by a rope at a 30o angle to the horizontal. The tension in the rope is 150 N, and the box is speeding up at 0.80 m/s2 (along the floor).

(a) How large is the frictional force of the floor on the box?


Homework Equations


Ffrict = uk*FN
FN = m*g - F*sin(theta)



The Attempt at a Solution


I used FN = m*g - F*sin(theta) to calculate the normal force, which is needed because it is a component of the equation to find frictional force. So, FN = (60)(9.8) - (150)sin30 => FN = 588 - 75 = 513
From here, I know that Ffrict = uk*FN, but the only thing I can plug in is 513 for FN, which still leaves me with two unknowns.

Can anyone be so kind as to tell me the next step? Am I missing something? Looking at the problem completely wrong?
 
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  • #2
What is the horizontal component of the 150 N force?

Then find the net horizontal force. That is equal to ma, which you can get from what you're given.

They don't give you μk, but you don't need it. You don't even need the normal force, although you got that just fine.
 
  • #3
SammyS said:
What is the horizontal component of the 150 N force?

Then find the net horizontal force. That is equal to ma, which you can get from what you're given.

They don't give you μk, but you don't need it. You don't even need the normal force, although you got that just fine.

The horizontal component would be 129.9 N, I believe. From using cos30 = adjacent/150.

As far as a net horizontal force, where does this come from? Is it just Fhorizontal = 60*0.80 = 48?
 
  • #4
revere21 said:
The horizontal component would be 129.9 N, I believe. From using cos30 = adjacent/150.

As far as a net horizontal force, where does this come from? Is it just Fhorizontal = 60*0.80 = 48?
Yes, 48 Newtons.
 
  • #5
SammyS said:
Yes, 48 Newtons.

So, I take the horizontal component of 129.9 N, and the net horizontal force of 48 N, and plug it into F - f = ma? Is that right? That would give me 129.9 - f = 48, which comes out to f = 81.9 N.

Can you give me one final hint and tell me if that is the right way to finish out this problem?
 
  • #6
Yes, that's right.
 
  • #7
SammyS said:
Yes, that's right.

Awesome. Thank you, yet again. You, sir, are a freshman-level physics wizard.
 

Related to Box dragged along floor at an angle, calculating normal and frictional force

1. What is the normal force in this scenario?

The normal force refers to the perpendicular force exerted by a surface on an object in contact with it. In the case of a box being dragged along the floor at an angle, the normal force would be equal to the weight of the box, since the floor is supporting its weight.

2. How do you calculate the normal force in this situation?

To calculate the normal force, you would need to use the formula: N = mg cos(theta), where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, and theta is the angle at which the box is being dragged. This formula takes into account the component of the weight of the box that is perpendicular to the surface.

3. What is the frictional force in this scenario?

The frictional force refers to the force that opposes the motion of an object in contact with a surface. In the case of a box being dragged along the floor at an angle, the frictional force would be the force of kinetic friction, since the box is in motion.

4. How do you calculate the frictional force in this situation?

To calculate the frictional force, you would need to use the formula: Ff = uN, where Ff is the frictional force, u is the coefficient of kinetic friction between the box and the floor, and N is the normal force (which can be calculated using the formula mentioned in question 2). The coefficient of kinetic friction is a measure of the roughness or smoothness of the surfaces in contact.

5. What other factors may affect the normal and frictional forces in this scenario?

Other factors that may affect the normal and frictional forces in this scenario include the weight of the box, the angle at which it is being dragged, the roughness of the surfaces in contact, and the presence of any external forces such as wind or inclines. Additionally, the values of the coefficient of kinetic friction and the acceleration due to gravity may vary depending on the surfaces and environment involved.

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