Question

A bulldozer drags a log weighing 500 n along a rough surface. the cable attached to the log makes an angle of 30 degrees with the ground. the coefficient of static friction between the log and the ground is 0.500. what is the minimum tension required in the cable in order for the log to begin to move?

Answer 1

T=Fn*u*sin30 = 500N * .5 * .5 = 125N

Answer 2

Answer:

The minimum tension required in the cable in order for the log to begin to move is $T=1154.7N$

Explanation:

In this problem we use Newton's equations, put a system of reference centered in the log (as a punctual mass), and analyse the forces in vertical y and horizontal x directions.

We get that

$N+Tsin(30)=W$

for y, where N is the normal force, T is what we want to know and W is given in the problem, for x we have

$Tcos(30)=\mu N=\mu W-\mu Tsin(30)$

where we have replaced N from the first equation, then we clear T

$T=\frac{\mu W}{\mu sin(30)-cos(30)}=\frac{0.5*500N}{0.5^2+\frac{\sqrt{3}}{2}}=1154.7N$

which is the answer to the problem.