Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
to provide a greater output force. In other words, we can lift a load without having to lift the full weight of the load. Prove using a mathematical relation that Class II levers provide mechanical advantage. You will need to use the rotational equilibrium condition. It will be helpful to draw a picture indicating the movement of the load. Also, to simplify we can assume the force is still perpendicular to by considering the case when the load just starts to move.
1 answer:
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
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Answer:
The mass of the heaviest box you will be able to move with this applied force = 61.4 kg
Explanation:
From the diagram attached, the forces acting on the box include the weight of the box, applied force on the box, normal reaction of the surface on the box and the Frictional force in the opposite direction to the applied force.
For the box to be able to move, the applied force must have a horizontal component that at least matches the Frictional force between the box and the surface. This is the force balance in the horizontal direction.
Resolving the applied force into horizontal and vertical components,
Fₓ = 750 cos 25° = 679.73 N
Fᵧ = 750 sin 25° = 316.96 N
Doing a force balance in the vertical axis,
N = (mg + 316.96)
Frictional force = μN = μ (mg + 316.96)
μ = 0.74, g = 9.8 m/s²
Frictional force = Fᵧ
μ (mg + 316.96) = 679.73
0.74(9.8m + 316.96) = 679.73
7.252m + 234.5504 = 679.73
7.252m = 679.73 - 234.5504 = 445.1796
m = (445.1796/7.252)
m = 61.4 kg
Hope this Helps!!!
To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.
PART A ) For the electrostatic force we have that is equal to
Here
q= Charge
E = Electric Force
PART B) Rearrange the expression F=ma for the acceleration
Here,
a = Acceleration
F = Force
m = Mass
Replacing,
PART C) Acceleration can be described as the speed change in an instant of time,
There is not then
Rearranging to find the velocity,