A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft.

A "scale" is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block E if the system is in equilibrium when s = 1.5 ft. 

Solution:

You can determine the weight of the suspended block 'E' BY USING COSINE LAW IN FORCE TRIANGLE.

applying the law of cos in the triangle, we will obtain the ff:
c = E
a = 10
b = 10
∝ = 132.84

where ∝ = 2θ - 180
∝ = 2(23.58) - 180
∝ = 132.84

substitute the value in cosine law where
c^2 = a^2 + b^2 - 2abcos∝

E^2 = 10^2 + 10^2 - 2(10)(10)cos132.84
E = 18.33

Determine thge magnitude and direction of the resultant force.

 

The curved rod lies in the x–y plane and has a radius of 3 m. If a force F = 80 N of acts at its end as shown, determine the moment of this force about point O.

 The curved rod lies in the x–y plane and has a radius of 3 m. If a force F = 80 N of acts at its end as shown, determine the moment of this force about point O.

The refrigerator has a weight of 200 lb and a center of gravity at G. Determine the force P required to move it. Will the refrigerator tip or slip? Take = 0.4.

The refrigerator has a weight of 200 lb and a center of gravity at G. Determine the force P required to move it. Will the refrigerator tip or slip? Take  = 0.4.

Solution

 tan (theta) = (meu_k) = 1.5/h; (meu_k) = 0.4.

0.4 = 1.5/h ; h = 3.75.

Where h = 4 (meu_k) = 1.5/4 = 0.375.

Then at (meu_k) = 0.4 the refrigeration is tips.

And at (meu_k) = 0.375 the refrigeration is slips.

Sum moments about other end equal to zero.

P*4-(200*1.5) = 0; p = 200*1.5/4; p = 75 lb tips.

The boy at D has a mass of 50 kg, a center of mass at G, and stands on a plank at the position shown. The plank is pin-supported at A and rests on a post at B. Neglecting the weight of the plank and post, determine the magnitude of force P his friend (?) at E must exert in order to pull out the post. Take B = 0.3 and C = 0.8.

The boy at D has a mass of 50 kg, a center of mass at G, and stands on a plank at the position shown. The plank is pin-supported at A and rests on a post at B. Neglecting the weight of the plank and post, determine the magnitude of force P his friend (?) at E must exert in order to pull out the post. Take _B = 0.3 and _C = 0.8.

 

Solution

Take vertical forces equal to zero, Ay + Ny = 490.5.
Take moment at A, then Ny*3=mg*2, By solving we get N=327.

Friction force = static friction coefficient * normal force.
f = 0.3 * 327 = 98.1;

Now take moment at c(post bottom),we get 

f(0.3+0.4) = Fcos30 * 0.3
By solving the above equation we get ;

F = 264.115 Newtons.

Express the force F1 in Cartesian vector form.

Express the force F₁ in Cartesian vector form.

Solution

 f = xi + yj + zk.

x is x component of force
y is y component of force
z is z component of force

By Considering the direction of the force
It is making 60 deg with positive y-axis and negative x-axis. and 45 deg with positive z-axis.
Cos theta(x) = x component of force/ magnitude of x.

Therefore x component = -(cos 60*400) = 200 since it is in negative direction
Similarly y component = cos 60 * 400 = 200.
z component = cos 45 * 400 = 283.
Therefore force in cartesian form is -200i+200j+283k.

The cable AO exerts a force on the top of the pole of F = {—120i — 90j — 80k} lb. If the cable has a length of 34 ft, determine the height z of the pole and the location (x,y) of its base.

The cable AO exerts a force on the top of the pole of F = {—120i — 90j — 80k} lb. If the cable has a length of 34 ft, determine the height z of the pole and the location (x,y) of its base.

Solution

 Here the force f has its value along x, y, z directions are 120, 90, 80.

So the resultant force is equal to 170.

So the angles are along x, y, z cos inverse of
(9120/170),(90/170),(80/170).

Now the length of the cable is 34.
So the components along,

x axis 34*(120/170) = 24.
y axis 34*(90/170) = 18.
z axis 34*(80/170) = 16.

The ball joint is subjected to the three forces shown. Find the magnitude of the resultant force.

The ball joint is subjected to the three forces shown. Find the magnitude of the resultant force.

Solution

 for

F1X = 1.25 (since f1 is parallel to x- axis)
F2y = 5(sin 30) = 2.5
F2z = 5(cos 30) = 4.330
F3y = 2(3/5) = 1.2
F3z = 2(4/5) = 1.6

summation along x = 1.25 = 1.25
summation along y = -2.5 + 1.2 = -1.3
summatiob along z = 4.330+1.6 = 5.93

R = SQUARE ROOT OF (fx+fy+fz)
R = SQUARE ROOT OF ((1.25)square)+ ((-1.3)square + ((5.93)square)
R = 6.20 KN

If F1 = F2 = 30lb, determine the angles and so that the resultant force is directed along the positive x axis and has a magnitude of FR = 20 lb.

If F₁ = F₂ = 30lb, determine the angles  and  so that the resultant force is directed along the positive x axis and has a magnitude of F_R = 20 lb.

 

Solution

It may be possible to solve this problem using Parallelogram Method cause in answer both θ and πare same,

R= √(P^2 + Q^2 +2P*Q*Cos(α))
and
Cos(α) = (R^2 - P^2 - Q^2)/(2*P*Q)
= (20^2 - 30^2 - 30^2)/(2*30*30)
= -0.778.

α = Cos ^-1(-0.778).
α = 141.06 °
As α =π+ θ = 2θ or 2π
Since in answer it is mention that both are equal.
θ = α/ 2,
= 141.06 / 2,
= 70.52°

The cord is attached between two walls. If it is 8 m long, determine the distance x to the point of attachment at B.

The cord is attached between two walls. If it is 8 m long, determine the distance x to the point of attachment at B.

 

 Solution

The diagram is not a right triangle, as it is in 3 dimensions and not on an x,y axis.

Correct approach:
A=0i+1j+4z.
B=?i+0j+2z.

Distance between 2 points in 3d:
sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2).

(x-0)^2 + (0-1)^2 + (2-4)^2.
x^2 + 1 + 4 = 8^2.
x^2=59.
x= sqrt59 = 7.68115.

Now if you check that you will see that it is correct, as opposed to the answer that you all tried to simplify and didn't bother to check your work

Determine the magnitude of the resultant force by adding the rectangular components of the three forces.

 Determine the magnitude of the resultant force by adding the rectangular components of the three forces.

Solution:

First, we must add/ sum up all the forces for x-components. That is...

Fx(Rx)= F1x + F2x + F3x (since F3 is parallel to y-axis,F3x=0)
= -50cos30 + 75cos45 + 50cos90
Fx(Rx)= 9.7317...N

Then, add/sum up all the forces for y-components.

Fy(Ry)= F1y + F2y + F3y
= 50sin30 + 75sin45 - 50sin90
Fy(Ry)= 28.03300...N

R=*squareroot of [(Ry)^2 + (Rx)^2]
=*suareroot of [(28.03300859 N)^2 + (9.7317384 N)^2]
R=29.67416895 N

θ= arctan [Ry/Rx]
= arctan [28.03300859/9.7317384]
θ= 70.86 degrees

Therefore, the answer is 29.67416895 N

Determine the resultant moment produced by forces F_BF B ​ and F_CF C ​ about point O. Express the result as a Cartesian vector.

 Determine the resultant moment produced by forces 

$F_B$ and $F_C$ about point O. Express the result as a Cartesian vector.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

To solve this question, we will be using part 1 and part 2 from the previous questions. Please refer to them if any part is confusing.

From part 1, we know that the moment produced by $F_B$ is:

$M_{O1}=\left\{-1800i\right\}N\cdot m$

 

From part 2, we know that the moment produced by $F_C$ is:

$M_{O2}=\left\{1080i+720j\right\}N\cdot m$

 

The resultant moment is then $M_{O1}+M_{O2}$:

$M_{resultant}=M_{O1}+M_{O2}$

$M_{resultant}=\left\{-1800i\right\}+\left\{1080i+720j\right\}$

$M_{resultant}=\left\{-720i+720j\right\}N\cdot m$

The friction at sleeve A can provide a maximum resisting moment of 125 N•m about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn.

 The friction at sleeve A can provide a maximum resisting moment of 125 N•m about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

Let us first express force F in Cartesian form. (Don’t remember?)

$F=\left\{-F\cos60^0i+F\cos60^0j+F\cos45^0k\right\}$

(simplify)

$F=\left\{-0.5F\,i+0.5F\,j+0.707F\,k\right\}$

 

We will now draw a position vector from A to B as follows:

Let us now calculate $r_{AB}$:

$r_{AB}=\left\{(-0.15-0)i+(0.3-0)j+(0.1-0)k\right\}$

$r_{AB}=\left\{-0.15i+0.3j+0.1k\right\}$

 

We can now calculate the moment along the x-axis. Remember that the unit vector for the x-axis is i.

$M_x=i\cdot r_{AB}\times F$

$M_x=\begin{bmatrix}1&0&0\\-0.15&0.3&0.1\\-0.5F&0.5F&0.707F\end{bmatrix}$

Taking the cross product gives us:

$M_x=0.1621F$

 

Substitute the maximum resisting moment (given to us in the question):

$125=0.1621F$

$F=771$ N