3D Statics problem

May 2019
7
0



Hello


Got a following problem.

300N Box on a frictionless inclined plane. The plane itself is in the Z-direction.
Box is held by a rope AB and a force in the direction Z. Determine the tension in the rope and the force F.
The answers to this problem are 104N and 215N


My question is where do i go wrong? the components, equations of equilibrium or both?


rope AB components that i calculated are following

ABx= 3.2m, ABy=-4.4m ABz=-3m
Box(x-dir)= 144N, Box(y-dir)= -300N

Equations of equilibriums would then go as follow:

\(\displaystyle sum Fx=(3.2/6.21)AB+144N=0\)
\(\displaystyle sum Fy= (-4.4/6.21)AB-300N=0\)
\(\displaystyle sum Fz= (-3/6.21)AB-F=0\)

144n comes from the incline itself, meaning that\(\displaystyle 300N*sin(36.87)*cos(36.87)\) and 300 is just mg in the direction -y
6.21 is the resultant of the force vector AB.

Personally i think the components are wrong but i would like to know for sure and i have no way of being sure of that myself so help is desperately needed. Thanks in advance
 

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Last edited:
Jan 2019
44
33
Forces acting on the box are force $F$ acting in the z-direction, weight, tension, and normal force, $N$, of the incline.

equation (1) $T_x - N_x = 0$
equation (2) $T_y + N_y = 300$

$\dfrac{T_x}{T_y} = \dfrac{8}{11} \implies T_x = \dfrac{8}{11} \cdot T_y$

$\dfrac{N_y}{N_x} = \dfrac{4}{3} \implies N_y = \dfrac{4}{3} \cdot N_x$

substituting for $N_y$ and multiplying all terms in equation (2) by $\dfrac{3}{4}$ yields

equation (1) $T_x - N_x = 0$
equation (2) $\dfrac{3}{4} \cdot T_y + N_x = 225$

combining the two equations yields

$T_x + \dfrac{3}{4} \cdot T_y = 225$

recall $T_x = \dfrac{8}{11} \cdot T_y$

$\dfrac{8}{11} \cdot T_y + \dfrac{3}{4} \cdot T_y = 225$

you should be able to solve for $T_y$ from this point, then $T_x$

finally, note $\dfrac{T_z}{T_y} = \dfrac{15}{22} \implies F = T_z = \dfrac{15}{22} \cdot T_y$

To three sig figs, you should get the values cited in your original post.
 
May 2019
7
0
Forces acting on the box are force $F$ acting in the z-direction, weight, tension, and normal force, $N$, of the incline.

equation (1) $T_x - N_x = 0$
equation (2) $T_y + N_y = 300$

$\dfrac{T_x}{T_y} = \dfrac{8}{11} \implies T_x = \dfrac{8}{11} \cdot T_y$

$\dfrac{N_y}{N_x} = \dfrac{4}{3} \implies N_y = \dfrac{4}{3} \cdot N_x$

substituting for $N_y$ and multiplying all terms in equation (2) by $\dfrac{3}{4}$ yields

equation (1) $T_x - N_x = 0$
equation (2) $\dfrac{3}{4} \cdot T_y + N_x = 225$

combining the two equations yields

$T_x + \dfrac{3}{4} \cdot T_y = 225$

recall $T_x = \dfrac{8}{11} \cdot T_y$

$\dfrac{8}{11} \cdot T_y + \dfrac{3}{4} \cdot T_y = 225$

you should be able to solve for $T_y$ from this point, then $T_x$

finally, note $\dfrac{T_z}{T_y} = \dfrac{15}{22} \implies F = T_z = \dfrac{15}{22} \cdot T_y$

To three sig figs, you should get the values cited in your original post.
Allrighty. i can folow the steps but few things pop up.
1. equation 2: \(\displaystyle Ty+Ny=300\) is the y component of the normal force \(\displaystyle 300*cos(36.87)\) but then taht gives me 240N and not 225. Basically im asking if you could explain how you got the components for the normal force?
 
May 2019
7
0
Thank you very much. Tell me if im wrong. What went wrong with my equations is 1. In Fx the Nx should have been 180N. 2. In Fy the Ny should have been 240N and also the 300N itself is acting against the y direction. Also what gave you the idea of doing the ratios at the beginning and doing them relative to each other instead of to the cable AB itself?
 
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