A heavy sled is being pulled by two people as shown in the figure. The coefficie

Question

A heavy sled is being pulled by two people as shown in the figure. The coefficient of static friction between the sled and the ground is As = 0.635, and the kinetic friction coefficient is = 0.403. The combined mass of the sled and its load is m = 336 kg. The ropes are separated by an angle = 23°, and they make an angle = 32.4° with thehorizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? Number If this rope tension is maintained after the sled starts moving, what is the sled's acceleration? Number m/s2

Explanation / Answer

if T is the minimum rope tension required to get the sled moving:

in the plane of the strings the resultant pulling tension will be 2Tcos. the mutually opposite Tsin from either string will cancel each other.

2Tcos = resultant tension at an angle of ° with the horizontal.

therefore horizontally the equivalent tension would be 2Tcoscos

this needs to equal static friction to begin movement.

=> 2Tcoscos = N -------------------------> (i)

vertically the equivalent tension would be 2Tcossin

N + 2Tcossin = mg = 336g

=> N = 336g - 2Tcossin

from equation (i) => 0.635[336(9.81) - 2Tcos23°sin32.4° ] = 2Tcos23°cos32.4°

=> (2093.06 – 0.6264T) = 1.554T

=> 2.18T = 2093.06

T = 960N

Once the sled is moving, the friction 'switches' to 'kinectic-mode'

also => F = 2Tcos23°cos32.4° - 0.403[336(9.81) - 2Tcos23°sin32.4°] = 336a

=> 1492.24 – 946.7 = 545.53 = 336a

=> a ~= 1.62 m/s

[Newton's II law: F = ma = F]

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