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Tricky vector question

  • Thread starter Avalanche
  • Start date
  • #1
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1. Homework Statement

A brick layer applies a force of 100 N to each of two handles of a wheelbarrow. Its mass is 20kg and it is loaded with 30 bricks,each of mass 1.5kg. The handles of the wheelbarrow are 30 degrees from the horizontal, and the coefficient of friction is 0.20. What initial acceleration is given the wheelbarrow?



2. Homework Equations

F = ma
F(g) = 9.8*m
F(f) = μ * F(n)




3. The Attempt at a Solution
I drew a free body diagram. The applied force is at the top left corner.

I found the vertical and horizontal components of the applied force
Vertical component = 100*sin 30° = 50 N
Horizontal component = 100*cos 30° = 86.60 N

I then multiplied both the horizontal and vertical component by 2 because its says the 100N force was applied to each handle

So vertical component = 100N and horizontal component = 173.2N

Total mass = 30*1.5+20 = 65kg

F(g) = 9.8*m
F(g) = 9.8*65 = 637 N

F(n) + vertical component = F(g) = 637N because the wheelbarrow is on the ground
F(n) + 100 = 637
F(n) = 537N

F(f) = μ * F(n)
F(f) = 0.20 * 537
F(f) = 107.4

Fnet = Horizontal component - F(f)
Fnet = 173.2 - 107.4
Fnet = 65.8

F = ma
65.8 = 65a
a = 1.01 m/s^2

But the answer is 0.71 m/s^2

What am I doing wrong?
 
Last edited:

Answers and Replies

  • #2
gneill
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There's something funny going on with your calculation of F(n); you indicate that you're adding the vertical component of the applied force, but it "doesn't take" -- the sum is still the weight of the wheelbarrow and contents.

That said, there's something fishy going on if the answer is supposed to be 0.71m/s^2. Was there a diagram with the problem? Perhaps the bricklayer's force is being applied in some "creative" fashion that's not coming across in the text.
 
  • #3
23
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There's something funny going on with your calculation of F(n); you indicate that you're adding the vertical component of the applied force, but it "doesn't take" -- the sum is still the weight of the wheelbarrow and contents.

That said, there's something fishy going on if the answer is supposed to be 0.71m/s^2. Was there a diagram with the problem? Perhaps the bricklayer's force is being applied in some "creative" fashion that's not coming across in the text.
There wasn't a diagram. That's usually my problem with these types of questions.

What do you mean by my calculation of F(n) "doesn't take?" My reasons for adding the vertical component and the normal force is that it should be equal to the force of gravity which is the weight of the wheelbarrow and contents
 
  • #4
gneill
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20,793
2,773
The vertical component of the applied force should add to the weight of the wheelbarrow (it's pushing down on it).
 
  • #5
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The vertical component of the applied force should add to the weight of the wheelbarrow (it's pushing down on it).
So you mean that the F(g) should be 637+100 = 737

So since F(n) = F(g), F(n) = 737

F(f) = μ * F(n)
F(f) = 0.2 * 737
F(f) = 147.4

Fnet = 173.2 - 147.4
Fnet = 25.8

F = ma
25.8 = 65a
a = 0.397 m/s^2

Like this? But the answer still is not 0.71 m/s^2
 
  • #6
gneill
Mentor
20,793
2,773
So you mean that the F(g) should be 637+100 = 737
Well, the normal force should be whatever net force is acting in the direction normal to the surface. In this case it's the sum of the weight of the wheelbarrow and the vertical force contributed by the brick layer's push.

So since F(n) = F(g), F(n) = 737

F(f) = μ * F(n)
F(f) = 0.2 * 737
F(f) = 147.4

Fnet = 173.2 - 147.4
Fnet = 25.8

F = ma
25.8 = 65a
a = 0.397 m/s^2

Like this? But the answer still is not 0.71 m/s^2
Yes, like that. And like I said, something looks fishy with either the given result or how the problem is being interpreted.
 
  • #7
you make it complicated, all you need is horizontal component and knowing that
Fnet=Fp - Ff
you can find Fnet and when you know Fnet=ma, then you can find a, because you know m :)
 
Last edited:

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