So to find the acceleration I did ( s - k)(g)= a I got the wrong answer btw! Ple

Question

So to find the acceleration I did (s - k)(g)= a

I got the wrong answer btw! Please help!

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 0.635, and the kinetic friction coefficient is 0.411. The combined mass of the sled and its load is m = 321 kg. The ropes are separated by an angle 22°, and they make an angle 32.4° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? minimum rope tension: 818.769 If this rope tension is maintained after the sled starts moving, what is the sled's acceleration? acceleration:2.19 m/s

Explanation / Answer

iven,

us = 0.635 ; uk = 0.411 ; m = 321 kg ; theta = 32.4 deg ; phi = 22 deg

W = mg = 321 x 9.8 = 3145.8 N = N

Ff1 = us N

Ff1 = 0.635 x 3145.8 = 1997.58 N

Let one perfon exerts force F

F' = 2 x F cos32.4 x cos22

F' = 1.57 F

In order for the sled to move

1.57 F = 1997.583

F = 1997.583/1.57 = 1272.35 N

T = 1997.583 N

Fnet = 1997.583 - 0.411 x 3145.8 = 704.66 N

ma = 704.6592

a = 704.6592/321 = 2.195 m/s^2

Hence, a = 2 m/s^2

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