A 10.0 kg mass is suspended on an incline by a rope connected to a wall

Question

A 10.0 kg mass is suspended on an incline by a rope connected to a wall. The coefficients of static and kinetic friction between the mass and the incline are 0.65 and 0.50, respectively. 

a) Draw a free-body diagram of the mass.

b) Determine the tension of the rope.

c) If the rope is cut, determine if the mass makes it to the bottom of the incline and if so how long it takes to reach the bottom. Assume the mass is 1.20 m from the bottom of the incline.

   

Answer 1

A free-body diagram of the mass should have gravity pointing directly downwards, the Normal force perpendicular to the incline, and the friction and tension forces parallel to the incline pointing up the slope.

The tension in the rope T should be equal to the component of gravity parallel to the incline F_gx, minus the friction force f, as F_gx=T+f for the system to be in equilibrium. Now, friction is defined as f = μ_sN= μ_smgcosθ, as the Normal force is equal to the component of gravity perpendicular to the incline, therefore:

T= F_gx-f = (mgsinθ)-(μ_smgcosθ)= mg(sinθ-μ_scosθ) then plug in the numbers.

The first part of c) can be checked with conservation of energy. If the work done by friction over 1.2m is greater than the gravitational potential energy, then the block will not make it to the bottom. Where work, w=Fd (force x distance)

fx > mgh = μ_kmgcosθx > mgxsinθ ⇒ μ_kcosθ > sinθ (h = xsinθ)

Making this an equality we can find the maximum angle before the block makes it to the bottom

μ_kcosθ = sinθ ⇒ tan^-1(μ_k)= θ = 26.565º. So if the angle of the incline is higher than this, then the block will slide off the incline.

Since the angle is not given to me I cannot say what the solution is, but if the angle is higher (which I assume it is) then the time can be found using a combination of conservation of energy and kinematics. Although it is odd other quantities are given but the angle is not, as θ does not cancel out.

We can use the equation v_f = v₀+at to solve for time.

v₀ is 0 because we're starting from rest.

a is found through Newton's Second Law:

ΣF=ma = F_gx-f = mgsinθ-μ_kmgcosθ = mg(sinθ-μ_kcosθ)

dividing by mass gives a:

a=g(sinθ-μ_kcosθ)

Now to find v_f through conservation of energy. The Gravitational potential energy minus the work done by friction will yield the leftover kinetic energy of the block a the bottom.

KE=1/2mv²=mgh-fx=mgxsinθ-μ_kmgxcosθ=mgx(sinθ-μ_kcosθ)

⇒ 1/2mv²=mgx(sinθ-μ_kcosθ) ⇒v=√[2gx(sinθ-μ_kcosθ)]

Back to our kinematic equation

v_f = v₀+at ⇒ v_f /a= t =√[2gx(sinθ-μ_kcosθ)]/g(sinθ-μ_kcosθ)

although this solution looks ugly it is a simple matter of plugging in numbers.

Using the range from 26.566° to 90° I've determined the solution to be between about 2 minutes to about half a second but again I can't say for sure without knowing a given angle.