Tension in ropes used to build the pyramids

[Priscilla]

Homework Statement

A mass M is hauled from ground level up an inclined plane that makes an angle 0 with the horizontal by means of a rope passing over a frictionless pulley. The mass is pulled along until it reaches a height of H. The building of the Egyptian pyramids and many other ancient construction jobs used such ramps. Logs were sometimes used as rollers to reduce friction.

a)Show that if the contact between the mass and ramp is frictionless, the work done by the tension in the rope (or equivalently, by the person hauling on the rope) is independent of the angle 0 .
b)Calculate the work done by the tension in the rope as a function of the ramp angle if the coefficient of kinetic friction between the mass and the surface is U_k.

Homework Equations

W=Fd
f_k = U_k*N
( N = normal force f_k = kinetic friction)

The Attempt at a Solution

6a, I have no idea how to prove it. Because I thought the tension is dependent of the angle. The angle is used to calculate the F_N, Fs...

6b
W=Fd d = H
N=mg cos 0
F=U_k mg cos 0
W=U_k mg cos 0 H

 

[Andrew Mason]

For a), what is the tension required just to keep the block moving (no acceleration)? (hint: think about what would it be if the ramp was perfectly horizontal; perfectly vertical - the answer is somewhere in between).


For b) do a free body diagram with the block moving up the ramp at constant speed. What must the tension in the rope be equal to if there is no acceleration? (hint: in order to work out the tension you have to determine the normal force, which is a function of the ramp angle.).

AM

 

[Andrew Mason]

(Private message).

For 6a) The way you tell me sound like it's dependent of the angle, but the question is asking to prove that the work done by tension is independent of the angle.

For 6b) The equations that I used are not correct?

You should use the thread so others can benefit.

Use the definition of work: $W = \vec{F}\cdot\vec{S}$. What is F (tension)? (hint: it is a function of $\theta$). What is the distance (hint: express in terms of h and $\theta$)? Multiply them together and see if the result involves $\theta$

b).Your equations are partially correct. You are missing the work done against the force of gravity.

You have to express the tension in terms of the other forces that balance the tension. Those forces are gravity and friction. Use a free body diagram and work them out.

AM

 

[Priscilla]

For a)
Tension on the inclined plane = Mg sin [0]
Distance of the inclined plane = H/ sin[0]
W = Fd = Td = (Mg sin[0]) (H/ sin[0]) = MgH

For b)
f_k = u_k N
N = Mg cos [0]
F = u_k Mg cos [0]
W = Fd
d = H/sin[0]
W = (u_k Mg cos [0] H)/ (sin[0])

Are they correct?

 

[Andrew Mason]

For a)
Tension on the inclined plane = Mg sin [0]
Distance of the inclined plane = H/ sin[0]
W = Fd = Td = (Mg sin[0]) (H/ sin[0]) = MgH

For b)
f_k = u_k N
N = Mg cos [0]
F = u_k Mg cos [0]
W = Fd
d = H/sin[0]
W = (u_k Mg cos [0] H)/ (sin[0])

Are they correct?

a) is correct. But not b). Compare your answer in b) to that in part a). Why would the work be LESS if you add friction?

AM

 

[Priscilla]

W = MgH + (u_k Mg cos [0] H)/ (sin[0])?

 

[Andrew Mason]

W = MgH + (u_k Mg cos [0] H)/ (sin[0])?

Looks right.

$$Tension = mg\sin\theta + \mu_k mg\cos\theta$$

$$W = F\cdot s =\frac{(mg\sin\theta + \mu_k mg\cos\theta)h}{\sin\theta}= mgh\left(1 + \frac{\mu_k}{\tan\theta\right)}$$

AM

 

[Priscilla]

Thanks!