A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 90 kg. The coefficient of static friction between the road and him is 0.86. The coefficient of kinetic friction between the road and him is 0.66. Find the tension in the cable.
Do we use static, or kinetic friction, or both to find tension? I am confused as to which one to consider.
Thanks!
He is moving, kinetic. The truck had to pull a little harder to get him moving.
F = .66 m g
We have to over-come static friction to get the motion started. We have to over-come kinetic friction to keep it going. We use kinetic friction because our system is already in motion.
M*g = 90 * 9.8 = 882 N. = Wt. of stuntman = Normal force, Fn.
Fk = u*Fn = 0.66 * 882 = 582 N. = Force of kinetic friction.
T-Fk = M*a.
T-882 = 90*0,
T = 882 N. = Tension in cable.
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Ah, the age-old question of friction! Well, in this particular scenario, since the stuntman is being pulled at a constant velocity, we can safely ignore the static friction. Now, cue the drumroll, please...
*drumroll*
The only type of friction we need to consider here is kinetic friction! It's the star of the show. So, wonderful question, but it looks like kinetic friction gets the spotlight this time. Now, let's find that tension in the cable!
To find the tension in the cable, we need to consider static friction, as the stuntman is being pulled at a constant velocity. When an object is at rest or in equilibrium, static friction comes into play. It prevents the object from moving until a certain force threshold is exceeded.
The formula to calculate static friction is:
F_static = µ_static * N
Where:
F_static is the static friction force
µ_static is the coefficient of static friction
N is the normal force
In this case, the normal force equals the weight of the stuntman, which can be calculated as:
N = m * g
Where:
m is the mass of the stuntman (90 kg)
g is the acceleration due to gravity (approximately 9.8 m/s²)
So, N = 90 kg * 9.8 m/s² = 882 N
Now, we can substitute the values into the static friction formula:
F_static = 0.86 * 882 N
Therefore, the static friction force is:
F_static = 759.72 N
Since the stuntman is being pulled at a constant velocity, the tension in the cable must match the opposing static friction force. So, the tension in the cable is also equal to 759.72 N.