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 s = 0.627, and the kinetic friction coefficient is k = 0.443. The combined mass of the sled and its load is m = 336 kg. The ropes are separated by an angle = 24°, and they make an angle = 32.1° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? If this rope tension is maintained after the sled starts moving, what is the sled's acceleration?

Explanation / Answer

suppose tension in rope is T.

then, equivalent force of both rope = 2Tcos(phi/2)

F = 2Tcos12

F = 1.956T
this force makees 32.1 degree from ground (above).

In vertical,

N + Fsin32.1 - mg = 0

N + 1.04T - (336 x 9.81) =0

N = 3296.16 - 1.04T

Max. friction = us. N = 0.627(3296.16 -1.04T)

In horizontal,

max. friction = Fcos32.1 (for minimum tension)

2066.69 - 0.652T = 1.952Tcos32.1

T = 896.35 N

----------------------------

once sled starts to move, now kinetic friction will work.

so friction = uk. N = 0.443 ( 3296.16 - 1.04T )

f = 1460.20 - 0.461T = 1460.20 - 0.461(896.35) = 1046.98 N

and in horizontal ,

Fcos32.1 - f = ma

1.956x896.35cos32.1 - 1046.98 = 336a

a = 1.30 m/s^2

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