Determining reaction forces and angle

[brostingy]

Homework Statement

An assembly is shown in Figure 1, GH, HK and HNL are three cables. The mass of box L is 350 kg,
a) Determine the mass of box K in order to make angle teta = 30 degrees
b) Determine the reaction forces at supports G and N.


i started by converting the 350kg to a force using F=ma, but stuck there...

help

digram attached to post

 

[droc22]

First, do FBD of the pulley that K is on.

$$\Sigma$$F_y=0=m_Lgsin30-m_kg

That solves m_k.

Reaction at G is x-component of the above FBD Gx = m_Lgcos30.

I'm sure you can figure out Rn now.

Hope that helps.

 

[brostingy]

ye thanks, i think i follow you, but just to clarify,

FDB is:

mass of L x gravity x sin30 - mass of K x gravity = 0

reaction at G:

G = mass of L x gravity x cos30

Rn:

should be:

y + x values...

sorry to sound dumb... :(

 

[droc22]

I won't give you the answers, but I'll give you the steps.

1. Draw FBD of Pulley K
2. Solve for x and y components to find mass of k and Rg. HINT: the magnitude (or hypotenuse) of the rope that's on 30 deg is the transferred weight of L
3. Draw FBD of Pulley L. You need a reaction force going up for Rn, and two forces (one on an angle of 30 deg-Also, remember 3rd Law) caused by the weight of Box L.
4. Split this diagram into x and y components and solve for Rnx and Rny, because the rest is know.

Do that make sense? Let me know what you get.

 

[paranormal]

To solve this problem, you will need to use the principles of static equilibrium. This means that the sum of all forces acting on the system must be equal to zero and the sum of all torques (rotational forces) must also be equal to zero.

To determine the mass of box K, we can use the fact that the tension in cable GH must be equal to the tension in cable HK since they are connected to the same point. This means that the vertical component of the force in cable GH must be equal to the vertical component of the force in cable HK. We can set up an equation to represent this:

Tsin(theta) = Tcos(theta)

Where T is the tension in both cables and theta is the angle of the cables (which is given as 30 degrees). We can rearrange this equation to solve for the mass of box K:

mK = (Tsin(theta))/g

Where g is the acceleration due to gravity (9.8 m/s^2). This will give you the mass of box K in kilograms.

To determine the reaction forces at supports G and N, we can use the principles of static equilibrium again. The sum of all forces in the vertical direction must be equal to zero, so we can set up an equation to represent this:

Fg + FN = mg

Where Fg is the force at support G, FN is the force at support N, and mg is the weight of box L. We can also use the fact that the sum of all torques must be equal to zero to solve for these forces. The torque due to the weight of box L (mg) will create a clockwise rotation, so the torque due to the reaction forces must create an equal and opposite counterclockwise rotation. This can be represented by the equation:

Fg(d) + FN(d) = mg(L/2)

Where d is the distance from the supports to the center of mass of box L and L is the length of the assembly. Solving these equations simultaneously will give you the reaction forces at supports G and N.

Remember, to solve problems like this, it is important to draw a clear and accurate free body diagram and use the principles of static equilibrium. I hope this helps!