physics tension problem?

Question

A 205kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30 degrees with respect to the horizontal. the coefficient of kinect friction between the log and the ramp is 0.900, and the log has an acceleration of 0.800 m/s^2. Find the tension of the rope.

I'm not sure how to work this problem..

Answer 1

m=205kg , z=30deg, u=0.9, a=0.8m/s^2

sum(F)=0
a is caused by T so ma and T are both up the ramp so they are positive
the force of gravity mgsinz and the force of
friction uN=umgcosz both point down the ramp
so they are negative, so
sum(F)=T + ma -mgsinz-umgcosz = 0

T= mgsinz + umgcosz - ma
=m(g(sinz + ucosz) - a)
= 2410N to 3sig.figs.

Answer 2

First...ID all the forces.

There is the pull P on the rope. There is the resistance to the pull due to friction (F) and our old friend gravity (W). And, as the log is accelerating, there is also f = ma, the net force on a mass (m) due to acceleration (a)

Second add them vectorially; thus...P - (F + W) = ma = f > 0 and set that sum equal to a net force f = ma (mass (m) times accleration (.8 m/sec^2). Since the only mass given is that of the log, we assume the rope, block and tackle, etc. are considered insignificant mass, m = 205 kg for the log.

As the rope is taut and not breaking, the pull P = T the tension you are looking for. Thus, rearrange the sum of forces so that P = ma + (F + W) and solve. As these are vectors, you need to change each one to scalars.

For example, as you did not give the direction of the acceleration, we assume it is parallel to the ramp. And given that the friction force is also parallel to the ramp, using the ramp as the reference is probably better than using the horizontal. That results, because the only force not parallel to the ramp in the P, W, F, and f mix of forces is W, the weight of the log.

But w = W sin(theta) is parallel to the ramp; so we put that into the sum of forces. Furthermore, N = W cos(theta) is the normal weight wrt the ramp and we need that in F = kN, the friction force acting along the ramp. Theta is the ramp incline angle = 30 deg. So now we can write:

P = ma + (W + F) = ma + w + F = ma + W sin(theta) + k W cos(theta) = ma + mg sin(theta) + k mg cos(theta) = m(a + g(sin(theta) + k cos(theta)), which are all scalar along the ramp and we can solve. Assume g = 9.81 m/sec^2 on Earth's surface and k = .9 the sliding friction coefficient. You have all the numbers you need; you can do the math.

But before you do that, stand back a minute and look at the equation. Ask yourself what does it mean. Well, it means, and this is the physics behind the equation, if we pull on a log, the tension in the rope will be equal to all the other forces on the log/rope system. Those other forces are the force due to acceleration of the mass, the force due to weight, and the force due to friction. And, moreover, the tension T is exactly equal to the pull P you put on the rope to bring the log up the ramp.

This latter is a really important physics concept you'll see over and over again in physics. It bears repeating, the pull on a taut rope is the tension on that rope. It has to be; otherwise the rope would not be taut or it would be breaking.

Answer 3

Physics Tension Problems And Solutions

Answer 4

you must identify all the forces involved, and how they cancel each other out

consider the force of gravity, pulling the log straight down, straight towards the center of the earth, the weight force

Fw = ma
Fw = (205kg)(9.8m/s^2)
Fw = 2009 N

The log is on a ramp, so this force will have horizontal and vertical components.
On any ramp, the horizontal component of the weight force can be found with Fwx = Fw(sin(angle)), and the vertical component is Fwy = Fw(cos(angle)).

Fwx = 2009N(sin(30)) = 1004.5 N
Fwy = 2009N(cos(30)) = 1739.8 N

using the horizontal component, which happens to be perpindicular to the ramps surface, we can determine how much the ramp pushes back; the normal force.

Fn = 1739.8 N

and THEN the force of friction, given with Ff = Fn*cof of friction

Ff = 1739.8N *.900 = 1565.9 N

the force of friction and the horizontal component of gravity are all the forces pulling the log DOWN the ramp.

Ff + Fwx
1565 . 9 N + 1004.5 N = 2570.4 N

the rope must have a tension greater than 2570.4 newtons, in order to get the block moving UP the ramp.

Answer 5

The weight mg of the log acts vertically down.

Resolving this into two components one perpendicular to the ramp and another along the ramp, we have,

mg cos 30 is perpendicular to the ramp pointing below, and mg sin 30 is along the plane pulling the mass down ward.

The reaction to mg cos 30 acts pointing up but inclined at angle 30 to the vertical.

The frictional force is 0.9 x mg cos 30. This friction is also acts along with mg sin 30 since the log moves up.

The Tension of the rope T acts up along the ramp.

The net force along the ramp is T - 0.9 x mg cos 30 - mg sin 30.

This force produces the acceleration on the mass m.

Therfore ,
T - 0.9 x mg cos 30 - mg sin 30 = ma.

T/m= a + 0.9 g cos 30 + g sin 30

T /m =13.34 N/ Kg.

T = 205 x 13.34 N
= 2734 .4 N.

Answer 6

Alright, hold on

you're going to have to draw a picture, a "free body diagram" if you havent already

there are four forces acting on the log. There is gravity, the normal force, friction, and the tension of the rope.

Gravity is directly down and is found by F = mg

so 205 x 9.81 is your gravititational force.

your normal force is at an angle of 30 degrees from straight up. your tension is 30 degrees from horizontal on the opposite side as the normal force. Friction is parallel and directly opposite your tension force

Now what you need to do is isolate what forces you have vertical (in the y or j-hat direction) and what forces you have horizontal (in the x or i-hat direction)

the forces for vertical and horizontal should each add up to zero, where a force down or to the left is negative and a force to the right or up is positive.

for vertical,

you have gravity which is negative, the y component of your normal force, the y component of your tension and the y component of your friction.

you should set up two equations which will have some unknowns in them. Use your sine, cos and tan knowledge to find the components. each must equal zero.

you should end up with two equations and two unknowns that you can substitute to solve for your unknowns which will give you the tension

good luck!