Question

A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground. The cable attached to the log makes an angle of 30° above the ground. The coefficient of static friction between the log and the ground is 0.50, and the coefficient of kinetic friction is 0.35. What minimum tension is required in the cable in order for the log to begin to slide?

Answer 1

Answer:

T= 224.01 N

Explanation:

in imminent motion we have to :

• The frictional force reaches its maximum value
• The system is in balance of forces

Data

W=  500 N :  weight of the log

μs = 0.5

μk = 0.35

α = 30°above the ground :  angle of the cable attached to the log

Newton's first law to the log:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

Forces acting on the log

T: cable tension for impending movement

N: normal force

W : weight

f: frictional force , f= μsN

We apply the formula (1)

∑Fx=0

Tx-f = 0

Tcosα-μsN=0

Tcos30°-0.5N=0 Equation (1)

∑Fy=0

N+Ty-W=0

N+Tsin30°-500=0

N= 500-Tsin30°  Equation (2)

We replace the value of N of the Equation  (2) in the equation (1)

Tcos30°-0.5(500-Tsin30°) = 0

Tcos30°+0.5Tsin30° = 0.5*500

T( cos30°+0.5*sin30°) = 250

(1.116) T = 250

T= 250/1.116

T= 224.01 N