Mechanics Assignment Help: Find Tension & Friction Coefficients

[jdstokes]

Hi,

I am going to submit one of those MasteringPhysics online assignments soon and I was wondering if anyone would be so kind as to check my answers.

http://session.masteringphysics.com/problemAsset/1010969/21/MLD_2l_1.jpg

The question is to find the tension in the upper and lower ropes when the system is at rest, and when it has acceleration a.

Case 1: The blocks are at rest. For M_1
T_1 - T_2 - M_1g = 0

For M_2
T_2 - M_2g = 0
so
T_2 = M_2g
and
T_1 = g(M_1 + M_2).

Case 2: The blocks are now accelerating upward. For M_1
T_1 - T_2 - M_1g = M_1a

For M_2
T_2 - M_2g = M_2a
so
T_2 = M_2(a+ g)
and
T_1 = (a + g)(M_1 + M_2).

http://session.masteringphysics.com/problemAsset/1010804/15/9374.jpg

For this question you need to find F in terms of T, given that each block has mass m. I used Newton's second law for each block to give
T_{AB} = ma (1)
T-T_{AB} = ma (2)
F-T = ma
combining (1) and (2) gives
ma = \frac{1}{2}T
so
F = \frac{3}{2}T.

http://session.masteringphysics.com/problemAsset/1010982/19/MLD_cm_7_a.jpg

A car rounding a frictionless banked curve with uniform speed v. Find the radius r of the turn.

I'm fairly sure the solution is r = \frac{v^2\cot\theta}{g}

Now \theta = 0 and there is a coefficient of static friction \mu between the road and the car's tyres. Find the mimimum value of the coefficient of static friction required to prevent the car from slipping. Assume that the car's speed is still v and that the radius of the curve is given by r. For this I reasoned that \frac{Mv^2}{r} is some multiple of the normal force Mg, so \mu_\mathrm{min} = \frac{v^2}{rg}.

http://session.masteringphysics.com/problemAsset/1010934/31/MFS_1l_3_v1_a.jpg

Find T_1.

Force balance in the horizontal and vertical directions gives

T_1\cos\theta_1 = T_2\cos\theta_2 (3)
and
T_1\sin\theta_1 + T_2\sin\theta_2 = mg. (4)
Eliminating T_2 from (4) gives
T_1\sin\theta_1 + T_1\cos\theta_1\tan\theta_2 = mg
so
T_1 = \frac{mg}{(\sin\theta_1 + \cos\theta_1\tan\theta_2)}.

http://session.masteringphysics.com/problemAsset/1010976/22/MLD_2l_2_v2_2_a.jpg

Find the ratio of the masses m_1/m_2.
For m_1
-m_1a = T - m_1g
T= m_1(g - a)
For m_2
m_2a = T - m_2g\sin\theta - \mu m_2g\cos\theta
m_2a = m_1(g - a) - m_2g\sin\theta - \mu m_2g\cos\theta
m_2(a + g\sin\theta + \mu g\cos\theta) = m_1(g - a)
m_1/m_2=\frac{a + g\sin\theta + \mu g\cos\theta}{g-a}

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.3×10^6 N, one at an angle 17.0 deg west of north, and the other at an angle 17.0 deg east of north, as they pull the tanker a distance 0.900 toward the north. What is the total work done by the two tugboats on the supertanker?

<br /> \begin{align*}<br /> W &amp; = \Sigma F_\mathrm{N}s \\<br /> &amp; = (2\times 1.3\times 10^6\thinspace\mathrm{N}\cos 17°)(900\thinspace<br /> \mathrm{m}) \\<br /> &amp; = 2.24\times 10^9\thinspace\mathrm{J}<br /> \end{align*}.

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle \alpha so that it reaches a stranded skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \mu_\mathrm{k}. Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier.

\frac{1}{2}mv^2 - \mu_\mathrm{k}mg\cos\alpha = mgh
v= \sqrt{2g(\mu_\mathrm{k}\cos\alpha + h)}.

 

[jdstokes]

I hate to bump, but I'm really intent on scoring full marks for this assignment :). If anyone has noticed any errors I'd greatly appreciate if they'd let me know.

Thanks.

James

 

[vincentchan]

The first one you are basically correct but,

Case 2: The blocks are now accelerating upward.

I think you meant the system instead of block, right?
The second one is correct, too.
For the third one, since you are so confident you did it right, I am not checking for you...
The forth one seems fine to me...
I can't read the rest of your problem... maybe the my browser has some problem reading letax... I'll check back later

good job..

 

[jdstokes]

I've tried splitting the post, maybe this will help you see the equations.

http://session.masteringphysics.com/problemAsset/1010982/19/MLD_cm_7_a.jpg

A car rounding a frictionless banked curve with uniform speed v. Find the radius r of the turn.

I'm fairly sure the solution is r = \frac{v^2\cot\theta}{g}

Now \theta = 0 and there is a coefficient of static friction \mu between the road and the car's tyres. Find the mimimum value of the coefficient of static friction required to prevent the car from slipping. Assume that the car's speed is still v and that the radius of the curve is given by r. For this I reasoned that \frac{Mv^2}{r} is some multiple of the normal force Mg, so \mu_\mathrm{min} = \frac{v^2}{rg}.

http://session.masteringphysics.com/problemAsset/1010934/31/MFS_1l_3_v1_a.jpg

Find T_1.

Force balance in the horizontal and vertical directions gives

T_1\cos\theta_1 = T_2\cos\theta_2 (3)
and
T_1\sin\theta_1 + T_2\sin\theta_2 = mg. (4)
Eliminating T_2 from (4) gives
T_1\sin\theta_1 + T_1\cos\theta_1\tan\theta_2 = mg
so
T_1 = \frac{mg}{(\sin\theta_1 + \cos\theta_1\tan\theta_2)}.

http://session.masteringphysics.com/problemAsset/1010976/22/MLD_2l_2_v2_2_a.jpg

Find the ratio of the masses m_1/m_2.
For m_1
-m_1a = T - m_1g
T= m_1(g - a)
For m_2
m_2a = T - m_2g\sin\theta - \mu m_2g\cos\theta
m_2a = m_1(g - a) - m_2g\sin\theta - \mu m_2g\cos\theta
m_2(a + g\sin\theta + \mu g\cos\theta) = m_1(g - a)
m_1/m_2=\frac{a + g\sin\theta + \mu g\cos\theta}{g-a}

 

[jdstokes]

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.3×10^6 N, one at an angle 17.0 deg west of north, and the other at an angle 17.0 deg east of north, as they pull the tanker a distance 0.900 toward the north. What is the total work done by the two tugboats on the supertanker?

<br /> \begin{align*}<br /> W &amp; = \Sigma F_\mathrm{N}s \\<br /> &amp; = (2\times 1.3\times 10^6\thinspace\mathrm{N}\cos 17°)(900\thinspace<br /> \mathrm{m}) \\<br /> &amp; = 2.24\times 10^9\thinspace\mathrm{J}<br /> \end{align*}.

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle \alpha so that it reaches a stranded skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \mu_\mathrm{k}. Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier.

\frac{1}{2}mv^2 - (\mu_\mathrm{k}mg\cos\alpha)(h\csc\alpha) = mgh
v= \sqrt{2gh(\mu_\mathrm{k}\cot\alpha + 1)}.

 

[dark_angel]

did u get any wrong?

 

[Optical_Man]

They all seem to be accurate to me.