Calculating tension/friction forces in a model

[wduk]

Hello

I am needing help trying to solve a problem.

I have an object at rest on a slope which is angled at 28 degrees, the weight of the object is 100 kg.

The only forces i need to worry about in this model, is the normal reaction, the friction, a tension string and the weight of the object. For my vectors i,j components i have angled them so i is parallel to the slope upwards and j is perpendicular to the slope.

The point of slipping for the object was uphill so i am assuming the friction vector acts down the slope.

So my vectors were:

N (Normal) = |N|j
S (the string attatched the object) = |S|i
F (Friction) = -|F|i 
W (Weight) = -100g cos(62)i - 100g sin(62) j  (not 100% sure if this is correct?)

By the way, as i am sure you know g = 9.8 ms^-2

So i am trying to find the magnitude of the friction, and then magnitude of the tension in the string.

What i have is that i know the object is at rest so using equilibrium i have:

N+S+F+W = 0

So for the i direction we have:

|S| - |F| - 100g cos(62) = 0
|N| + 100g sin(62) = 0

So i can solve N easily but i can't find the value of F or S unless i know the value of one of them right? So have i missed something, how can i find their magnitudes?

 

[bboyson]

Response to question re tension and friction

The Force along the slope without friction is Wsin28 where W=100kg and therefore that is what the tension would have to be to keep the weight from sliding. The normal force is give by Wcos28 and the friction is determined by the product of the Normal Force N and the friction coefficient,u. So if you want to determine the net force available to move the block it would be WSin28-uN. Does that help

Bert Boyson

 

[wduk]

So the Frictional force is equal to the normal reaction's magnitude multiplied by the friction coefficient? I thought that only applied to a flat surface not one on an incline?

So basically F is : μ|N|


And the tension is S = F + 100 g cos(62)

Is that what you are saying?

 

[bboyson]

Friction

The Force along the slope without friction is Wsin28 where W=100kg and therefore that is what the tension would have to be to keep the weight from sliding. The normal force is give by Wcos28 and the friction is determined by the product of the Normal Force N and the friction coefficient,u. So if you want to determine the net force available to move the block it would be WSin28-uN. Does that help

Bert Boyson

Re your question about the tension, as long as there is a force normal to the direction of travel there is a frictional resistance even on a slope. In the case quoted the normal force is the 100 kg weight x the cosine of 28 degrees. Notice that as the slope angle approaches zero, the cosine approaches 1.