A mountain climber (m1= 75kg) hangs vertically from a cliff, tied to her friend standing on an inclined plane at 22° degrees above the horizontal. If the force of friction between the rope and the plane is 72 N,
What minimum mass of her friend will stop her from falling?
If the friend’s mass is 86 kg, calculate the acceleration of both climbers.
m1*g-72-massfriend*g*sin22=0 If I am viewing it in my mind correctly., and assuming no friction holding the friend.
To find the minimum mass of her friend that will stop her from falling, we need to consider the forces acting on the climber and her friend.
Let's draw a free-body diagram for the climber.
1. Label the downward force due to gravity acting on the climber as mg, where m is the mass of the climber (75 kg) and g is the acceleration due to gravity (typically 9.8 m/s^2).
2. Label the tension force in the rope as T.
3. Label the frictional force between the rope and the inclined plane as f (72 N).
Now, let's break down the weight of the climber into its components along the inclined plane.
1. The component of gravity acting parallel to the inclined plane is mg*sin(22°). This component acts to pull the climber down the incline.
2. The component of gravity acting perpendicular to the inclined plane is mg*cos(22°). This component does not contribute to the motion along the incline.
The force of friction, f, opposes the downhill motion of the climber and can be expressed as f = μ*N, where μ is the coefficient of friction and N is the normal force.
The normal force, N, is equal to mg*cos(22°). Therefore, f = μ*mg*cos(22°).
Since we want to find the minimum mass of her friend to stop her from falling, we need to determine the maximum possible frictional force that can be exerted by the rope. This occurs when the climber is on the verge of slipping.
In this case, the force of friction equals the maximum possible force of friction. Therefore, we have f = μ*mg*cos(22°) = 72 N.
Now, we can solve for the coefficient of friction:
μ*mg*cos(22°) = 72 N
μ * (75 kg) * (9.8 m/s^2) * cos(22°) = 72 N
Solving the equation, we find:
μ ≈ 0.294
Now, let's calculate the minimum mass of her friend that will stop her from falling. The net force acting on the climber in the horizontal direction is given by:
Net force = T - μ*mg*sin(22°)
For the climber to remain stationary (no acceleration), the net force in the horizontal direction must be zero.
Therefore, T = μ*mg*sin(22°)
Substituting the known values, we get:
T = (0.294) * (75 kg) * (9.8 m/s^2) * sin(22°)
Solving this equation, we find that T ≈ 95.4 N.
Since the tension in the rope is equal to the normal force exerted by the friend, and the normal force is equal to the weight of the friend, we have:
Weight of friend = T = 95.4 N
Therefore, the minimum mass of the friend required to stop the climber from falling is given by:
Weight of friend = mass of friend * g
95.4 N = mass of friend * (9.8 m/s^2)
Solving for the mass of the friend, we find:
mass of friend = 9.75 kg (approximately)
Now, let's move to the second part of the question: calculating the acceleration of both climbers given that the mass of the friend is 86 kg.
Since the two climbers are connected by a rope, they will have the same acceleration. We can use Newton's second law to find this acceleration.
The net force acting on each climber is given by the sum of the forces acting on them:
Net force = T - mg*sin(22°)
Since the climbers have the same acceleration, we have:
T - mg*sin(22°) = (m_climber + m_friend) * a
Substituting the known values, we get:
(0.294)*(75 kg)*(9.8 m/s^2) - (75 kg)*(9.8 m/s^2)*sin(22°) = (75 kg + 86 kg)*a
Simplifying the equation, we find:
20.573 - 16.5138 = 161 kg * a
Solving for the acceleration, we get:
a = (20.573 - 16.5138) / 161 kg
a ≈ 0.025 m/s^2
Therefore, the acceleration of both climbers is approximately 0.025 m/s^2.