A straight path is inclined at an angle of 15 degree to the horizontal. A loaded skip of total mass 1500 kg is at rest on the path and is attached to a wall at the top of the path by a rope. The rope is taut and parallel to a line of greatest slope of the path, as shown in the diagram. Calculate the normal and rictional components of the contact force exerted on the skip by the path when the tension in the rope is 2000 N.
After the rope is cut the skip is on the point of slipping down the path. Calculate the coefficient of the friction between the skip and the path
0.268
To calculate the normal and frictional components of the contact force exerted on the skip by the path, we need to break down the forces acting on the skip.
1. Normal Force (N): The normal force is the perpendicular component of the contact force exerted by the path on the skip. It acts in a direction perpendicular to the surface of the path, counteracting the weight of the skip. The magnitude of the normal force can be calculated using the formula:
N = m * g * cos(θ)
where m is the mass of the skip (1500 kg), g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of inclination (15 degrees).
Plugging in the values, we get:
N = 1500 * 9.8 * cos(15°)
2. Frictional Force (Ff): The frictional force is the parallel component of the contact force exerted by the path on the skip. It acts in the direction opposite to the intended motion, resisting the skip from sliding down the path. The magnitude of the frictional force can be calculated using the formula:
Ff = µ * N
where µ is the coefficient of friction between the skip and the path.
Plugging in the value of N from the previous step, we get:
Ff = µ * (1500 * 9.8 * cos(15°))
To find the coefficient of friction (µ) when the rope is cut and the skip is on the point of slipping down the path, we can set the frictional force equal to the maximum force of static friction:
Ff = µ * N = µ * (1500 * 9.8 * cos(15°))
Since the skip is on the point of slipping, the maximum static friction is equal to the tension in the rope (2000 N):
Ff = 2000
Now we can solve for µ:
2000 = µ * (1500 * 9.8 * cos(15°))
Rearranging the equation:
µ = 2000 / (1500 * 9.8 * cos(15°))
Simplifying this expression will give you the numerical value of the coefficient of friction between the skip and the path.
1+1=2
2+2=4