How to solve this kinetic friction problem? KINDLY PLEASE SHOW ME ON HOW TO DO IT.. THANK YOU VERY MUCH?

Question

A man pulls a trunk of mass M across a level floor using a massless rope that makes an angle with the horizontal. The coefficient of frciton between trunk and floor is Mk. What rope tension is required to move the trunk at constant velocity.

Answer 1

Constant Velocity means, there is no acceleration, a = 0 It means, you must not use F = ma. Just use Equilibrium considerations.



Here's the figure:



...............Ty

........W'....↑..........⁄..T

.........↓ ....|........⁄

╔═════╗|....⁄...Ø

║'..''..M...║|⁄-----------→Tx

╚═════╝←-----Ff

▒▒▒↑▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

.......N



These are the forces acting on the trunk.



Fx = T cos Ø

Fy = T sin Ø



If you take summation of Forces along the Vertical direction (along the Y-axis),

forces going upwards will be positive(+) and forces going downwards negative(-)



∑Fv = 0

0 = N + Ty - W -------> could be also written ( N + Ty -W = 0)

N = W - Ty



The relation of N and Ff is given as



Ff = Mk(N).... where Mk = coefficient of friction

Substitute the value of N



Ff = Mk (W - Ty)

Ff =Mk ( mg - Tsin Ø)



Now if you take summation of Forces along the Horizontal (along the X-axis) forces going to the right positive(+) and those going to the left negative (-)



∑Fн = 0

0 = - Ff + Tx

Ff = Tx

Ff = Tcos Ø



Ff = Ff

Tcos Ø = Mk ( mg - Tsin Ø)

...............= Mk(mg) - Mk(Tsin Ø)

Tcos Ø + Mk(Tsin Ø) = Mk(mg)

T(cos Ø + Mksin Ø) = Mk(mg)

Answer 2

You would need to know the angle the rope makes with the horizontal, which we would call θ (theta). Then, if the rope has tension T, the vertical component will be T*sinθ and the horizontal component will be T*cosθ. I will use m for mass and μk (mu sub-k) for the coefficient of kinetic friction, to avoid confusion with the use of capital M.



Basically, the component of tension in the horizontal direction needs to be equal to the force of kinetic friction. This will put the trunk into equilibrium, letting it be pulled at a constant velocity. The force of kinetic friction is equal to the normal force times the coefficient of kinetic friction. The normal force is equal to the weight of the trunk less the vertical component of tension (since it pulls the trunk upwards). So the normal force is mg - sinθ and the force of friction is (mg - sinθ)*μk. Setting this equal to the horizontal component of tension, we have (mg - sinθ)*μk = cosθ.

Answer 3

CONSTANT velocity? that means ( if no friction) F=0

If with friction :

F cos angle - Mk X M X g = M X a

since a=0,

F cos angle = Mk X M X g

So the force needed is equal to the friction