A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
1 answer:
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
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<h2><u>We have</u>,</h2>
- Initial velocity (u) = 0 m/s
- Time taken (t) = 2.9s
- Acceleration due to gravity (g) = + 10 m/s² [Down]
- Final velocity (v)
- Height (h)
→ v = u + gt
→ v = 0 + 10(2.9)
→ v = 29 m/s … ( Ans )
And,
→ h = ut + ½gt²
→ h = 0(2.9) + ½ × 10 × (2.9)²
→ h = 5 × 8.41
→ h = 42.05 m … ( Ans )
1. 12.75 J
Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is
where
F = 15 N is the magnitude of the force
d = 85 cm = 0.85 m is the displacement of the cart
Substituting in the formula, we get
2. 10.6 N
In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).
Therefore, the work done is
However, in this case the displacement is
d = 120 cm = 1.20 m
Therefore, the magnitude of the force in this case is
