Model the forces as constant over a short interval of time

[shiri]

The figure below shows a worker poling a boat-a very efficient mode of transportation-across a shallow lake. He pushes parallel to the length of the light pole, exerting on the bottom of the lake a force of 225 N. Assume the pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of 30.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 360 kg.

(a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force.

(b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.400 s after the moment described.
I got the answer for (a), 3.34e3 N. I calculate:
[Force of gravity] = [mass][gravity] = [360kg][9.81m/s^2] = 3531.6 N
[Normal force] = [Force exerted on bottom of the lake]*cos30 = 225N*cos30 = 194.9 N
[Magnitude of the force] = [Force of gravity]-[Normal force] = 3531.6 N-194.9 N = 3337N
But I couldn't find the answer for (b) What I got so far is to calculate the:

(Net force) = (Horizontal drag force) + [(Force exerted on the bottom of the lake)*(cos
60)] = -47.5N + (225*cos 60) = 65.0N

then find accerleration, a

a = F/m = 65.0N/360kg = 0.180556m/s^2

then find v(0.400)

v = at = 0.180556m/s^2 * 0.400s = 0.07222m/s

then add both velocities

0.007222m/s + 0.857m/s = 0.9292m/s = 9.29e-1m/s

and I don't understand why it's a wrong answer. Can you figure out why?

 

[Doc Al]

But I couldn't find the answer for (b) What I got so far is to calculate the:

(Net force) = (Horizontal drag force) + [(Force exerted on the bottom of the lake)*(cos
70)] = -47.5N + (225*cos 70) = 29.4545N

Why cos(70)? You want the horizontal component of the pole force.

 

[shiri]

Why cos(70)? You want the horizontal component of the pole force.

yes, I want the horizontal component of the pole force

 

[Doc Al]

But where did the 70 degrees come from? The only angle given was 30 degrees.

 

[shiri]

But where did the 70 degrees come from? The only angle given was 30 degrees.

oh...my bad...the angle given is 30 degrees and to look for the horizontal component should be 60 degrees. sorry

***i recalculate on the first reply of this topic

 

[Doc Al]

Good. That's better.

Your revised speed looks OK to me. Why do you say its wrong?

 

[shiri]

Good. That's better.

Your revised speed looks OK to me. Why do you say its wrong?

well when I submit the answer, 9.29e-1, it comes out like this

"Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

but I never rounded any values while I'm calculating. So I was wondering did I do calculations right or it is completely wrong?

 

[Doc Al]

I don't see anything wrong with your method or calculation. I get the same numbers. (Sometimes those online systems can be flaky.)

 

[shiri]

I don't see anything wrong with your method or calculation. I get the same numbers. (Sometimes those online systems can be flaky.)

So even your own work and calculations come out with a same answer?

 

[Doc Al]

Yes.

 

[shiri]

Yes.

Do you know any way to make this question correct so it doesn't show up this reason below:

"Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

I only got two submissions left