Since there is no tension, there are only two relevant forces acting on the block: friction and gravity. Since the block is motionless, we can also write:

\(\displaystyle F_{net} = 0\)

\(\displaystyle F_g - F_f = 0\)

\(\displaystyle F_g = F_f\)

Substitute the expressions for these two forces:

\(\displaystyle mgsin(\theta) = \mu_s gcos(\theta)\)

Canceling out mass and gravitational acceleration, and rearranging for the coefficient of static friction, we get:

\(\displaystyle \mu_s = \frac{sin(\theta)}{cos(\theta)}=tan(\theta)\)

\(\displaystyle \mu_s = 0.36\)

There are three relevant forces acting on the block in this situtation: friction, gravity, and tension. We can use Newton's second law to express the system:

\(\displaystyle F_{net}=ma\)

\(\displaystyle F_g -F_f - T = ma\)

Substituting expressions in for the forces, we get:

\(\displaystyle mgsin(\theta) - \mu_kmgcos(\theta)-T = ma\)

Canceling out mass and rearranging to solve for tension, we get:

\(\displaystyle T = \frac{gsin(\theta)-\mu_kgcos(\theta)}{a}\)

We have values for each variable, allowing us to solve:

\(\displaystyle T = \frac{(10\frac{m}{s^2})sin(30^{\circ})-(0.55)(10\frac{m}{s^2})cos(30^{\circ})}{2\frac{m}{s^2}}\)

\(\displaystyle T = 0.12 N\)

In order to find the mass of block 2, we're going to need to calculate a few other things, such as the tension in the rope.

To begin with, we'll need to identify the various forces on our free-body diagram. To do this, we will begin with block 1 and use a rotated coordinate system to simplify things. In such a system, the x-axis will run parallel to the surface of the ramp, while the y-axis will be perpendicular to the ramp's surface, as shown below:

Now we can identify the forces acting on block 1. Along the rotated y-axis, the force of gravity acting on the block is equal to \(\displaystyle mg\cos(\theta)\), and the force of the ramp on the block is just the normal force, \(\displaystyle N\). Since block 1 is not moving in the y direction, we can set these two forces equal to each other.

\(\displaystyle N=m_{1}gcos(\theta )\)

Now, considering the forces acting along the rotated x-axis, we have a force pointing downwards equal to \(\displaystyle m_{1}g\sin\left ( \theta \right )\). Pointing upwards, we have the tension force \(\displaystyle T\) and we also have the frictional force, \(\displaystyle F_{k}\).

The formula for calculating the force due to kinetic friction is:

\(\displaystyle F_{k}=\mu _{k}N\)

Since we have already determined what the normal force is, we can substitute that expression into the above equation to obtain:

\(\displaystyle F_{k}=\mu _{k}m_{1}gcos(\theta )\)

Now, we can write an expression for the net force acting upon block 1 in the x direction:

\(\displaystyle F_{1(net)}=m_{1}a=m_{1}gsin\left ( \theta \right )-T-\mu _{k}m_{1}gcos(\theta )\)

Rearrange the above expression to solve for tension.

\(\displaystyle T=m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a\)

So far, we have only been looking at block 1. Now let's turn our attention to block 2 and see what forces are acting on it. In the downward direction we have the weight of the block due to gravity, which is equal to \(\displaystyle m_{2}g\). In the upward direction, as we can see in the diagram, we have the tension of the rope, \(\displaystyle T\). We need to write an expression that tells us the net force acting upon block 2.

\(\displaystyle F_{2(net)}=m_{2}a=T-m_{2}g\)

Since we calculated the expression for tension from the information regarding block 1, we can plug that expression into the above equation in order to obtain:

\(\displaystyle m_{2}a=[m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a]-m_{2}g\)

Now rearrange to solve for the mass of block 2.

\(\displaystyle m_{2}a+m_{2}g=m_{2}(a+g)=m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a\)

\(\displaystyle m_{2}=\frac{[m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a]}{(a+g)}\)

Then plugging in values, we can finally calculate block 2's mass:

\(\displaystyle m_{2}=\frac{[(12kg*9.8\frac{m}{s^{2}}* sin(30^{o}))-(0.18* 12kg* 9.8\frac{m}{s^{2}}* cos(30^{o}))-(12kg* 1.5\: \frac{m}{s^{2}})]}{(1.5 \frac{m}{s^{2}}+9.8 \frac{m}{s^{2}})}\)

\(\displaystyle m_{2}=2.28kg\)

Since the gravitational force must be cancelled by the tension force, as the ball is experiencing no acceleration, and no other forces are being applied to it:

\(\displaystyle F_{grav}=F_{tension}=mg\)

\(\displaystyle F_{grav}=10kg*9.8 \frac{m}{s^2}=98N= F_{tension}\)

The block has three forces on it: the force of tension, the force of gravity, and the force from Bruce. The force of gravity is:

\(\displaystyle F_{grav}= mg\)

\(\displaystyle F_{grav}=10kg*10\frac{m}{s^2}=100N\)

The force from Bruce plus the force of tension has to equal gravity (since Bruce's force and tension are up while gravity is down) so the block is in equilibrium. 

\(\displaystyle F_{grav}=F_{Bruce}+F_{T}\)

\(\displaystyle 100N=3N+F_{T}\)

\(\displaystyle F_{T}=97N\)

\(\displaystyle F_{total}=F_1+F_2+...\)

\(\displaystyle ma=F_{rope}+F_{gravity}\)

\(\displaystyle ma=T+mg\)

Plug in values:

\(\displaystyle 1500*1=T+1500*-9.8\)

Solve for \(\displaystyle T\)

\(\displaystyle T=16.2kN\)

\(\displaystyle F_{cent}=T=m\frac{v^2}{r}\)

The velocity of the earth is \(\displaystyle \frac{\left(2\pi*149.6*10^9meters\right)}{1 year}\)

Convert to \(\displaystyle \frac{meters}{second}\):

\(\displaystyle \frac{9.40*10^{11}meters}{year}*\frac{1 year}{365.25 days}*\frac{1 day}{24 hours}*\frac{1 hour}{3600 seconds}=\frac{2.99*10^4m}{s}\)

Plug in values:

\(\displaystyle F_{cent}=T=5.97*10^{24}kg\frac{\left (\frac{2.99*10^4m}{s}\right)^2}{149.6*10^9m}\)

\(\displaystyle T=3.57*10^{22}N\)

Here is a (simplistic) diagram of the elevator. There are three forces acting on this object. There is it's own weight due to gravity \(\displaystyle mg\); the tension in the rope holding it up, which we will call \(\displaystyle T\); and there is an external force \(\displaystyle ma\) due to the fact that the elevator is accelerating upward. Since the mass is fixed, it has zero velocity with respect to the elevator, and therefore the net force on the mass is 0 by Newton's 2nd law. The net force is given by \(\displaystyle T+ma-mg=0\). Solving this equation for \(\displaystyle T\) and plugging in the values gives us:

\(\displaystyle T=mg-ma=(4kg)(9.8\frac{m}{s^2})-(4kg)(2\frac{m}{s^2})=31.2N\)

Since we want the magnitude of the tension, and not the value of the vector, we omit the minus sign. Therefore the tension in the rope is \(\displaystyle 31.2N\), as desired.

For this question, we're presented with a scenario in which an object of a given mass is hanging to a rope. That rope is being pulled on, causing the object to accelerate upwards at a certain rate. We're then asked to calculate the tension that results in the rope.

In order to answer this question, it's best if we approach this by examining a force diagram. Since there are no pertinent forces occurring in the x direction, the only concern we have is in the y direction.

One of the forces acting on the object is the downward force of gravity. Another force is the upward tension caused by the rope. Hence, we know what force components contribute to the net force. Furthermore, we're told that as it is pulled up, the object accelerates at a rate of \(\displaystyle 8\: \frac{m}{s^{2}}\). Since this is the net acceleration in the y direction, we can determine the net force in the y direction.

\(\displaystyle \Sigma F_{y}=T-mg=ma\)

Next, we can rearrange the above terms in order to isolate the term for tension.

\(\displaystyle T=mg+ma=m(g+a)\)

Finally, if we plug in the values that we know, we can calculate our answer.

\(\displaystyle T=(25\: kg)(8\: \frac{m}{s^{2}}+9.8\: \frac{m}{s^{2}})\)

\(\displaystyle T=445\: N\)

The masses will each provide an equal force to each other through the rope. Thus, there will be no net force and no net acceleration.