A stubborn, 120 {\rm kg} mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 800 {\rm N}. The coefficients of friction between the mule and the ground are \mu _{\rm s} = 0.80 and \mu _{\rm k} = 0.50. Is the farmer able to move the mule?
What is your thinking on this?
forcefriction=mu*mg
pulling force-forcefriction=mass*accleration
To determine whether the farmer is able to move the mule, we can analyze the forces acting on the mule.
The maximum force that the farmer can exert is given as 800 N. This is the force that the farmer exerts on the mule through the rope.
There are two different types of friction that affect the movement of objects on a surface: static friction and kinetic friction.
- Static friction occurs when an object is at rest and prevents it from moving. The magnitude of static friction can be given by the product of the coefficient of static friction (\mu_s) and the normal force (N). In this case, the normal force is equal to the weight of the mule, which is the mass (m) multiplied by the acceleration due to gravity (g). Therefore, the static friction force (F_{\text{static}}) can be calculated as:
F_{\text{static}} = \mu_s \cdot N
= \mu_s \cdot m \cdot g
- If the force exerted by the farmer exceeds the maximum static friction force, the object will start to move. Once the object is in motion, kinetic friction comes into play. The magnitude of kinetic friction can be given by the product of the coefficient of kinetic friction (\mu_k) and the normal force (N). Therefore, the kinetic friction force (F_{\text{kinetic}}) can be calculated as:
F_{\text{kinetic}} = \mu_k \cdot N
= \mu_k \cdot m \cdot g
In this case, the weight of the mule can be calculated as:
N = m \cdot g
= 120 \, \text{kg} \times 9.8 \, \text{m/s}^2
Now, we can plug in the given values to calculate the static friction force and kinetic friction force.