A bench is made of a 3 meter long plank with a mass of 12kg and 2 sawhorses(each 0.2m from an end).There are two boxes on the bench. A 15kg box is 0.8 meter from the left hand side. A second box is 0.5 m from the right hand side. What would the mass of the second box have to be for the normal force exerted by the right sawhorse to be 3 times larger than that exerter by the left sawhorse?
You need three equations to solve for the unknown mass M and the forces at each sawhorse, F1 and F2
One of the equations is F2 = 3 F1
Another equation can be
F1 + F2 = 4 F1 = (12 + 15 + M)g
The third equation should be a moment equation. I suggest you take it about the point of contact of the sawhorse on the right.
F1*2.6 - 15g*2.0 - 12g*1.3 - Mg*0.3 = 0
You should draw yourself a figure of the loaded plank to see where the lever arm length coefficients came from.
Now you have two equations in two unknowns, M and F1. Solve for M. The g factor can be canceled out when solving for M.
To find the mass of the second box, we can start by considering the forces acting on the bench.
The weight of the plank itself can be calculated using the formula:
Weight = mass × acceleration due to gravity
Weight = 12 kg × 9.8 m/s² (acceleration due to gravity)
Weight = 117.6 N
Next, let's label the distances from the left side of the bench to the different objects:
- Distance from the left side to the left sawhorse (L) = 0.2 m
- Distance from the left side to the right sawhorse (R) = 3 m - 0.2 m = 2.8 m
- Distance from the left side to the 15 kg box (A) = 0.8 m
- Distance from the left side to the second box (B) = 3 m - 0.5 m = 2.5 m
Now we can consider the forces acting on the bench:
1. The weight of the plank itself (117.6 N) acts at the center of the plank (1.5 m from the left end).
2. The normal force exerted by the left sawhorse (F₁) acts upward at a distance of 0.2 m from the left end.
3. The normal force exerted by the right sawhorse (F₂) acts upward at a distance of 2.8 m from the left end.
4. The weight of the 15 kg box (15 kg × 9.8 m/s²) acts downward at a distance of 0.8 m from the left end.
5. The weight of the second box (m₂ × 9.8 m/s²) acts downward at a distance of 2.5 m from the left end.
Now, we can set up equilibrium equations for the forces in the vertical direction:
ΣFᵥ = 0
Downward forces (weight of the plank, weight of the 15 kg box, and weight of the second box) are positive, while upward forces (normal forces from the sawhorses) are negative.
117.6 N - 15 kg × 9.8 m/s² - m₂ × 9.8 m/s² - F₁ - F₂ = 0
Next, we can set up an equation to represent the condition where the normal force exerted by the right sawhorse (F₂) is three times larger than that exerted by the left sawhorse (F₁):
F₂ = 3F₁
We can substitute this equation into the equilibrium equation:
117.6 N - 15 kg × 9.8 m/s² - m₂ × 9.8 m/s² - F₁ - 3F₁ = 0
Simplifying the equation:
117.6 N - 15 kg × 9.8 m/s² - m₂ × 9.8 m/s² - 4F₁ = 0
Now, let's substitute the distance values into the equation to calculate F₁:
F₁ = (117.6 N - 15 kg × 9.8 m/s²) / 4
F₁ = (117.6 N - 147 N) / 4
F₁ = -29.4 N / 4
F₁ ≈ -7.35 N
Note: Negative value indicates that the force is directed in the opposite direction (upward instead of downward).
Now we can substitute this value of F₁ back into the previous equilibrium equation:
117.6 N - 15 kg × 9.8 m/s² - m₂ × 9.8 m/s² - (-7.35 N) - 3(-7.35 N) = 0
Simplifying the equation further:
117.6 N - 147 N - m₂ × 9.8 m/s² + 7.35 N + 22.05 N = 0
117.6 N - 147 N + 29.4 N + 22.05 N = m₂ × 9.8 m/s²
21.05 N = m₂ × 9.8 m/s²
Now, we can solve for the mass of the second box, m₂:
m₂ = 21.05 N / 9.8 m/s²
m₂ ≈ 2.15 kg
Therefore, the mass of the second box would have to be approximately 2.15 kg for the normal force exerted by the right sawhorse to be three times larger than that exerted by the left sawhorse.