F, F. F, F, Figure 5.32 (a) What force does each side of the nutcracker shown in Figure 5.32 exert on a nut 3.10 cm from the pivot? The hand exerts a perpendicular force of 52.5 N on each side at a distance of 10.0 cm from the pivot. (b) What is the total force on the nut? (No calculations are necessary for this part.) What keeps the nut from slipping out?
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
![### Figure 5.32
**Description:**
The figure shows a nutcracker, which is an example of a simple lever system. The nutcracker is depicted with two handles, and a nut is held between the jaws. There are annotated forces, \( F_i \) and \( F_o \), acting on different parts of the system.
**Diagram Explanation:**
- **Forces:**
- \( F_i \) is the input force applied by the hand on each side of the nutcracker.
- \( F_o \) is the force exerted on the nut due to the lever action.
- **Measurements:**
- The hand applies a perpendicular force of 52.5 N at a distance of 10.0 cm from the pivot.
- The nut is located 3.10 cm from the pivot.
**Questions:**
(a) What force does each side of the nutcracker exert on a nut 3.10 cm from the pivot? The hand exerts a perpendicular force of 52.5 N on each side at a distance of 10.0 cm from the pivot.
\[ \text{Answer:} \quad \_\_\_\_\_\_ \, \text{N} \]
(b) What is the total force on the nut? (No calculations are necessary for this part.)
\[ \text{Answer:} \quad \_\_\_\_\_\_ \, \text{N} \]
**Discussion:**
- **Leverage and Force Exertion:** The nutcracker utilizes the principle of leverage, allowing a smaller force to generate a larger force to crack the nut. The mechanical advantage depends on the ratio of the distances from the pivot.
- **Friction:** Friction between the nut and the nutcracker jaws is essential in preventing the nut from slipping out during the operation.
This provides a practical illustration of how levers work and how force is amplified in simple machines.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcca7b3a9-28ae-495e-82df-ff344407dffc%2F7ae306f9-b0d5-467b-bf62-cb6736179913%2F6ub2f5j_processed.jpeg&w=3840&q=75)
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