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Avalanche
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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?
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?
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