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

AI Thread Summary
A 60 kg box is pulled at a 30-degree angle with a tension of 150 N, resulting in an acceleration of 0.80 m/s². The normal force was calculated as 513 N, but the frictional force can be determined without needing the coefficient of friction. The horizontal component of the tension force is approximately 129.9 N. The net horizontal force is calculated as 48 N, leading to the equation 129.9 N - f = 48 N, resulting in a frictional force of 81.9 N. The problem-solving approach was confirmed as correct by participants in the discussion.
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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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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.
 
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?
 
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.
 
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?
 
Yes, that's right.
 
SammyS said:
Yes, that's right.

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