Is it possible to solve for kinetic friction without given theta?

[joe426]

Homework Statement

A block of mass " lying on a ramp tilted at an
angle  with the horizontal is connected to
another block of mass # by a light string passing
over a frictionless pulley as shown. If the block
slides up the ramp with a constant velocity ),
what is the coefficient of kinetic friction $%?

Homework Equations

f=ma

The Attempt at a Solution

ƩF_1xx = m₁a
F_t - μ_kF_n = m₁a

ƩF_1y = m₁a_1y
F_n - mgcosθ = m₁a_1y


I can figure out the F_t from the other mass. But where I am stuck is I have 3 unknowns and 2 equations. I can't find a way to solve for μ_k when I don't have θ.

Thanks for the help

 

[voko]

I cannot see the second mass in the equations.

 

[voko]

I cannot see the second mass in the equations.

 

[joe426]

ƩF_2y = m₂a
F_T - m₂g = m₂a

I solve this for F_t and plug that into the tension force on mass1

 

[voko]

So what are the three unknowns you have?

 

[joe426]

So what are the three unknowns you have?

μk,Fn, and θ

 

[voko]

It is not clear from the initial statement of the problem what is given and what is not. The initial statement sounds to me as if the angle were on an equal footing with masses which I assume you consider given. Could you copy the problem exactly?

 

[joe426]

What I have in the original post is the whole question but I have attached a picture of the problem for clarity.

 

[voko]

I think you can consider the angle as a given.

 

[lendav_rott]

The assignment says that the block moves with a constant velocity - Newton's 1st law says that the end result Force affecting the mass is 0 if the body is still or is moving with a constant velocity and without changing direction - acceleration is 0.

This is the part that I am doubting at the moment -is it correct to assume the mass m₂ is pulling the mass m₁ by its weight? If so, the weight is m₂g - and as the block on the ramp isn t accelerating, the mass on the other side also isn't accelerating.

So the resulting force to the block on the ramp is m₂g.
The frictional force or force of friction or ..:/ is F_h = μm₁gcosθ

The ramp directed gravitational pull is F = m₁g sinθ

The resulting force is m₂g + (-m₁g sinθ - μm₁gcosθ) <- these are vectors - i don't know how to denote vectors :(
the weight of m₂ has to come level with both the friction and the gravitational pull.
So after a bit of simplification:
(m₂g)² + (-m₁g(sinθ + μcosθ))² = 0

E: I meant you must know the value of all 3 variables - even the assignment given agrees with me.