You are pushing an 80.0 N wheelbarrow as shown in (Figure 1). You lift upward on the handle of the wheelbarrow so that the only
point of contact between the wheelbarrow and the ground is at the front wheel. Assume the distances are as shown in the figure, where 0.50 m is the horizontal distance from the center of the wheel to the center of gravity of the wheelbarrow. The center of gravity of the dirt in the wheelbarrow is assumed to also be a horizontal distance of 0.50 m from the center of the wheel. Suppose that the maximum total upward force that you can apply to the wheelbarrow handles is 75 lb.
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Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
