Question

A man is hauling a box of mass 100kg up a 35 degree incline by a rope attatched to the top of the box. If the rope makes an angle of 20 degrees to the incline, and the coefficient of friction between the box and the incline is 0.65, calculate the force applied by the man to keep the box moving at a constant speed.

Answer 1

Weight of box= mg=100*9.8N=?

Answer 2

\[980N\]

Answer 3

This weight will b acting just downward. Right?

Answer 4

Yes

Answer 5

Should I type what I did?

Answer 6

This weight 980 N has 2 perpendicular component

Answer 7

Yes. One perpendicular to the plane, and one parallel to the plane (right)?

Answer 8

Okb. U can type wt u did. Plz!!!

Answer 9

:)

Answer 10

Ya. Right

Answer 11

Ok, first i found what gravity was:
\[Fg=m\times g\]\[Fg=100*9.8\]\[Fg=980N\]

Answer 12

Ok

Answer 13

I then find the force pushing down, parallel to the plane
\[Fp=Fg\times sin\theta\]\[Fp=980N*sin(35)\]\[Fp\approx561.9N\]

Answer 14

I then split the force of the pull into its \(x\) and \(y\) components

Answer 15

\[Fa_x=Fa\times \sin(20)\]\[Fa_y=Fa*\cos(20)\]

Answer 16

No

Answer 17

Oh, K. What do I do then

Answer 18

Again, to get frictional force, you need to find the N, which is mgsin@ or mgsin35, that is 100*9.8*cos35=802.8N. Frictional force or R is $N or 0.65*802.8=521.8N, this is the frictional force due to the slopes friction.
Next, take account of the driving force, which is the one acting at 20degrees, so the Fd=Tension*cos20 or Tcos20 .
Finally, we can now use the Newtons formula
Fd -R =ma. Given that the speed is constant, then a= 0, rewriting the formula as Fd=R
Take note that the forces acting downward apart from the friction one is also that of weight, which is mgsin35 or 100*9.8*sin35=562.1N
Then, we shall have Tcos20= 562.1 + 521.8. And T=1153N.

Answer 19

U split pulling force along the plane nd perpendicular to the plane

Answer 20

@shamim, isn't that what I did? After I said I was gonna split it (I never splt Fp)

Answer 21

It is just that the tension acts parallel to the plane, when tension is multiplied by cos20, which is what I made as driving force.

Answer 22

U r right @Ahsome

Answer 23

Phew. Got to do something. Can I explain the rest in a bit?

Answer 24

Ok. Keep going

Answer 25

I think u hv 2 unequal but oposit forces

Answer 26

U can minus frm bigger

Answer 27

Me?

Answer 28

Why u r silent @Ahsome

Answer 29

@shamim, I said I needed to do something for a little bit. Can I come back and we can discuss in a little bit (dinner)

Answer 30

Ok

Answer 31

But, didn't my explanation also facilitated a bit?

Answer 32

Bon Appetite anyway. Over here is still 2:09pm.

Answer 33

Which country u r? @Hoslos

Answer 34

Back You guys still here, @shamim and @Hoslos?

Answer 35

Ok. I m here

Answer 36

Ok. So, we have the following things:
\[Fg=980N\]
\[Fa_y=Fa\times \sin(20)\]\[Fa_x=Fa\times \cos(20)\]
\[Fp=561.N\]

Answer 37

Now, last thing we need to do is find \(Fn\). Normally, \(Fn\) is equal to \(Fg\times \cos(\theta)\), but in this situation part of the pull force pushes the object up. Whenthe object goes up, the \(Fn\) becomes smaller. In the end:

\[Fn=Fg\times \cos\theta - y\]\[Fn=980\times cos(35)-y\]

Answer 38

Ok. So, we have the following things:
\[Fg=980N\]
\[Fa_y=Fa\times \sin(20)\]\[Fa_x=Fa\times \cos(20)\]
\[Fp=561.N\]
\[Fn=980\times cos(35)-y\]

Answer 39

We know the equation that \(Fr=Fn\times µ\) Where Fr is friction, Fn is normaly and µ is co-efficient of friction

Answer 40

I am from southern Africa. What about you @shamim?

Answer 41

Ok. So, we have the following things:
\[Fg=980N\]
\[Fa_y=Fa\times \sin(20)\]\[Fa_x=Fa\times \cos(20)\]
\[Fp=561.N\]
\[Fn=980\times cos(35)-y\]
\[Fr=Fn\times µ\]

Answer 42

Welcome back @Hoslos :D

Answer 43

Let me continue...

Answer 44

Yes!

Answer 45

Since we know that the object is moving at a constant velocty, then whatever is pushing up has to be equal to pushiing down. The force pushing up is the x value of the pulling force. The forces pushing down (the plane) are Friction and the part of gravity parallel to the plane

Answer 46

\[Fr+Fp=x\]

Answer 47

When you do Fgsin35 it is the force of weight acting parallel to the slope, which is correct.

Answer 48

Yes

Answer 49

Correct for forces acting down the plane.

Answer 50

But the force that acts up the plane is Tcos20, because it is said to act at a given angle.

Answer 51

We know that \[Fr=Fn\times µ\]. So lets put that into the equation:
\[Fr+Fp=x\]\[Fn×µ+Fp=x\]

Answer 52

Eventually Tcos20= friction + force of weght parallel to plane.

Answer 53

Exactly

Answer 54

\[Fn×µ+Fp=x\]

Answer 55

Ya

Answer 56

Lets slowly sub everything we know

Answer 57

Right!!!

Answer 58

So 561 is of weight and 803 is friction, right?

Answer 59

We don't know friction yet (I haven't showed that yet ;) ) Put yes, 561 is the force parallel to the plane

Answer 60

I mean friction is 522, right?

Answer 61

I am not sure

Answer 62

To get this, you first find force to the y-plane, being mgcos35 or 100*9.8*cos35=802.8, which is the contact force. This is the first step when going for friction. Are you ok up to this point?

Answer 63

Yes, but you need to take away the y component of the pulling force, since that does part of the work for Fn, so Fn is smaller

Answer 64

Correct @Ahsome

Answer 65

Hmm, I am seeing that we different ways of solving this problem. But I believe mine is simpler. The way I learnt was that when you have an inclined plane and want to find frictional force, you first find the y-plane force which is mgcos$ to get the normal contact force and after that we do frictional force=Mu*N , where Mu is the coefficient of fricion and N is the contact force.

Answer 66

Exactly. That works if no force moves in the same direction as Fn

Answer 67

But the thing is, the pulling force does move in that direction (the y component)

Answer 68

So you need to take that into consideration aswell

Answer 69

So, my point after finding the 802.8N, was to then base from the formula R=MuN or 802.8*0.65=521.8N - the frictional force.

Answer 70

But remember, the normal force is smaller than 802.8

Answer 71

Right @shamim? (I hope I am)

Answer 72

Right@Ahsome

Answer 73

Yes, because you have to subtract with the Tsin20.

Answer 74

Yup. But you don't know T?

Answer 75

That is what thewuestion wants.

Answer 76

Yup. So how could you assume that Fn is 801

Answer 77

What I am thinking is that we make that normal contact force subtracting the Tsin20 as one equation and we multiply with the Mu then add with the forceof weight parallel to plane. This will be the equation of the forces acting downwards.
And we equate to Tcos20, which is the force acting upwards, making as find T the subject of the formula, because we will have an equation.