A straight horizontal rope is towing a 5.52 kg toboggan across frictionless ice. A 2.76 kg radio is on top of the toboggan with a coefficient of static friction of 1.50 between the toboggan and radio. In order to prevent the radio from slipping, what is the largest force of tension for the rope?
To find the largest force of tension for the rope, we need to consider the forces acting on the toboggan and the radio.
Let's analyze the forces first:
1) Tension force (T): This is the force applied by the rope to pull the toboggan.
2) Weight of the toboggan (Wtob): This force is equal to the mass of the toboggan (5.52 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
3) Weight of the radio (Wr): This force is equal to the mass of the radio (2.76 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
4) Frictional force (Ffr): This force depends on the coefficient of static friction (μ) between the toboggan and the radio. It is given by the equation Ffr = μ × (Wtob + Wr).
Since we want to prevent the radio from slipping, the frictional force (Ffr) must be equal to or greater than the force that would cause the radio to start sliding. This force is the product of the coefficient of static friction and the normal force acting on the radio (which is equal to the weight of the radio, Wr).
Mathematically, this can be represented as Ffr ≥ μ × Wr.
To find the largest force of tension, we need to find the maximum static frictional force and then add the weight of the toboggan and the radio.
The maximum static frictional force (Fmax) can be found using the equation Fmax = μ × (Wtob + Wr).
Plugging in the given values:
μ = 1.50 (coefficient of static friction)
Wtob = 5.52 kg × 9.8 m/s^2 (weight of the toboggan)
Wr = 2.76 kg × 9.8 m/s^2 (weight of the radio)
Calculating:
Fmax = 1.50 × (5.52 kg × 9.8 m/s^2 + 2.76 kg × 9.8 m/s^2)
Simplifying:
Fmax = 1.50 × (53.296 kgm/s^2 + 27.048 kgm/s^2)
Fmax = 1.50 × 80.344 kgm/s^2
Fmax ≈ 120.516 kgm/s^2
Finally, to find the largest force of tension for the rope, we add the weight of the toboggan and the radio to the maximum static frictional force:
Largest force of tension = Fmax + Wtob + Wr
Largest force of tension ≈ 120.516 kgm/s^2 + (5.52 kg × 9.8 m/s^2) + (2.76 kg × 9.8 m/s^2)