I'm doing some last minute studying for my Physics exam
A bulldozer drags a log weighing 5.00 x 102 N along a rough surface. The cable attached to the log makes an angle of 30.0° with the ground. The coefficient of static friction between the log and the ground is 0.500. What minimum tension is required in the cable in order for the log to begin to move?
I got this wrong on a quiz, and don't understand how to do it.
I've seen this, http://answers.yahoo.com/question/index?…
but I don't understand it at all.
A bulldozer drags a log weighing 5.00 x 102 N along a rough surface. The cable attached to the log makes an angle of 30.0° with the ground. The coefficient of static friction between the log and the ground is 0.500. What minimum tension is required in the cable in order for the log to begin to move?
I got this wrong on a quiz, and don't understand how to do it.
I've seen this, http://answers.yahoo.com/question/index?…
but I don't understand it at all.
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OK, I'll paste in the text from the link you provided and then insert explanations with a > to introduce what I inserted..
Let:
T = tension on the cable.
N = normal force
Ff = force of friction
N = 500 - T sin30
>N = 500 - 0.5T
>N is the force pressing the log to the ground. The normal force is less than the log's weight because the tension in the cable actually lifts while pulling.
Ff = 0.5 N = 0.5(500 - 0.5T) = 250 - 0.25T
> The formula for friction is Ff = mu_s*N where mu_s is the coefficient of static friction.
Ff = Tcos 30
>We know it starts moving at some amount of tension. The horizontal force is the horizontal component of T: Tcos30 This is the lowest value of T that is successful.
250 -0.25T = 0.866T
>There are 2 expressions for Ff above. What he did here is set the 2 expressions equal to each other. So then he just solves for T.
1.116T = 250
T = 224.014N
Let:
T = tension on the cable.
N = normal force
Ff = force of friction
N = 500 - T sin30
>N = 500 - 0.5T
>N is the force pressing the log to the ground. The normal force is less than the log's weight because the tension in the cable actually lifts while pulling.
Ff = 0.5 N = 0.5(500 - 0.5T) = 250 - 0.25T
> The formula for friction is Ff = mu_s*N where mu_s is the coefficient of static friction.
Ff = Tcos 30
>We know it starts moving at some amount of tension. The horizontal force is the horizontal component of T: Tcos30 This is the lowest value of T that is successful.
250 -0.25T = 0.866T
>There are 2 expressions for Ff above. What he did here is set the 2 expressions equal to each other. So then he just solves for T.
1.116T = 250
T = 224.014N