friction for both wedge surfaces is 0.30 and if the coefficient of static friction between the block and the horizontal surface is 0.60, determine the least force P justed by the 5° wedge under the action of the force P. If the coefficient of static The horizontal position of the 500-kg rectangular block of concrete is ad- required to move the block.
Example 6.6
a)The direction of the IM (impending motion) of the Block is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
b)The direction of the IM (impending motion) of the Wedge is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
c)Consider the FBD of the Block:
The magnitude of the angle that the R2 force is off the horizontal is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
d)Consider the FBD of the Block:
The direction of N2 (normal force on Surface-2) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
e)Consider the FBD of the Block:
The direction of F2 (friction force on Surface-2) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
f)Consider the FBD of the Block:
The magnitude of the angle that the R3 force is off the vertical is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
g)Consider the FBD of the Block:
The direction of N3 (normal force on Surface-3) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
h)Consider the FBD of the Block:
The direction of F3 (friction force on Surface-3) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
i)Consider the FBD of the Wedge:
The magnitude of the angle that the R2 force is off the horizontal is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
j)Consider the FBD of the Wedge:
The direction of N2 (normal force on Surface-2) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
k)Consider the FBD of the Wedge:
The direction of F2 (friction force on Surface-2) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
l)Consider the FBD of the Wedge:
The magnitude of the angle that the R1 force is off the horizontal is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
m)Consider the FBD of the Wedge:
The direction of N1 (normal force on Surface-1) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
n)Consider the FBD of the Wedge:
The direction of F1 (friction force on Surface-1) is _____degrees.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
o)The magnitude of R2 (reaction force on Surface-2) is _____N.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
p)The magnitude of N2 (normal force on Surface-2) is _____N.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
q)The magnitude of F2 (friction force on Surface-2) is _____N.
NOTE:
- If the author directly provides this information, enter the number of Sig Figs as the author provides.
![**Sample Problem 6/6**
**Problem Statement:**
The horizontal position of the 500-kg rectangular block of concrete is adjusted by the 5° wedge under the action of the force P. If the coefficient of static friction for both wedge surfaces is 0.30 and if the coefficient of static friction between the block and the horizontal surface is 0.60, determine the least force P required to move the block.
**Solution:**
The free-body diagrams of the wedge and the block are drawn with the reactions \( R_1, R_2, \) and \( R_3 \) inclined with respect to their normals by the amount of the friction angles for impending motion. The friction angle for limiting static friction is given by \( \phi = \tan^{-1} \mu \). Each of the two friction angles is computed and shown on the diagram.
We start our vector diagram expressing the equilibrium of the block at a convenient point A and draw the only known vector, the weight W of the block. Next we add \( R_3, \) whose 31.0° inclination from the vertical is now known. The force vector \( R_2, \) whose 16.7° inclination from the horizontal is also known, must close the polygon for equilibrium. Thus, point B on the lower polygon is determined by the intersection of the known directions of \( R_3 \) and \( -R_2, \) and their magnitudes become known.
For the wedge we draw \( R_2 \), which is now known, and add \( R_1 \), whose direction is known. The directions of \( R_1 \) and P intersect at C, thus giving us the solution for the magnitude of P.
**Algebraic Solution:**
The simplest choice of reference axes for calculation purposes is, for the block, in the direction a-a normal to \( R_3 \) and, for the wedge, in the direction b-b normal to \( R_1 \). The angle between \( R_2 \) and the a-direction is 16.7° + 31.0° = 47.7°. Thus, for the block
\[
[\Sigma F = 0] \quad 500(9.81) \sin 31.0° - R_2 \cos 47.7° = 0
\]
\[
R_2 = 375](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcffe6fb8-a8b1-4738-951c-b41ccaeecf4e%2F32c525e9-f826-4600-96fb-54562bb76c2c%2F6b2hhrq_processed.jpeg&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 5 steps with 2 images
