A straight horizontal rope is towing a 6.13 kg toboggan across frictionless ice. A 3.06 kg radio is on top of the toboggan with a coefficient of static friction of 2.07 between the toboggan and radio. In order to prevent the radio from slipping, what is the largest force of tension for the rope
If the toboggan accelerates very fast, the radio will slip relative to the toboggan (like pulling a tablecloth from under plates - the plates stay behind if the table-cloth's acceleration is big enough).
Think of the radio separately:
Weight: W = mg
Normal reaction: Fn = W = mg
Limiting frictional force = μFn = μmg
The radio's horizontal acceleration is caused by the horizontal frictional force (as this is the only horizontal force on the radio).
The maximum frictional force that can act on the radio is the limiting frictional force, μmg. This produces an acceleration (using F = ma):
a = F/m = μmg/m = μg = 0.450*9.81 = 4.4145m/s²
The system's total mass M = 6.16+3.08 = 9.24kg
The maximum acceleration is produced by the maximum tension T. Since T is the only external horizontal force on the system:
T = Ma = 9.24*4.4145 = 40.8N
how did you get the 0.450 ?
To find the largest force of tension that can be applied to the rope without the radio slipping, we need to consider the maximum static friction force between the toboggan and the radio.
The maximum static friction force (F_static_max) can be found using the formula:
F_static_max = µ * N
where µ is the coefficient of static friction and N is the normal force.
The normal force (N) acting on the radio is equal to the weight of the radio (m_radio * g), where m_radio is the mass of the radio and g is the acceleration due to gravity (9.8 m/s^2).
Given:
m_radio = 3.06 kg (mass of the radio)
g = 9.8 m/s^2
N = m_radio * g
= 3.06 kg * 9.8 m/s^2
≈ 29.96 N
Substituting the values into the formula:
F_static_max = 2.07 * 29.96 N
≈ 61.85 N
Therefore, the largest force of tension that can be applied to the rope to prevent the radio from slipping is approximately 61.85 N.