Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
Answer:
(B) (length)/(time³)
Explanation
The equation x = ½ at² + bt³ has to be dimensionally correct. In other words the term bt³ and ½ at² must have units of change of position = length.
We solve in order to find the dimension of b:
[x]=[b]*[t]³
length=[b]*time³
[b]=length/time³
Answer:
The pressure corresponding to the absolute zero temperature is 0.997atm.
Explanation:
To solve this question, you draw a straight vertical line with the boiling point temperature and pressure on top of the line and the freezing point temperature and pressure on the lower part. The absolute temperature somewhere in the middle of the line with the pressure to be obtained.
So, we have;
0- (-19) / 100 - (-19) = P - 0.9267 / 1.366 - 0.9267
19 / 119 = P - 0.9267 / 0.4393
Cross multiply, we have
19 * 0.4393 = 119(P -0.9267)
8.3467 = 119P - 110.2773
119P = 118.624
P = 0.997 atm
So at 0°C, the pressure of the thermometer is 0.997atm.
Answer:
0.243
Explanation:
<u>Step 1: </u> Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
<u>Step 2:</u> calculate the torque required to raise the load
![T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BR%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B5%2B%5Cpi%280.09%29%2825%29%7D%7B%5Cpi%2825%29-0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (9.66 + 6.75)N.m
= 16.41 N.m
<u>Step 3:</u> calculate the torque required to lower the load
![T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BL%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B%5Cpi%280.09%29%2825%29%20-5%7D%7B%5Cpi%2825%29%2B0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (1.64 + 6.75)N.m
= 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = 
= 
= 0.243
<h3><u>Answer;</u></h3>
33.9 pounds
<h3><u>Explanation</u>;</h3>
In order for the ladder to be in equilibrium, the net torque should be equal to zero. Therefore, the torque in the opposite directions should equal each other:
Clockwise torque = Counter clockwise torque
Torque is the product of the applied force and the distance between that force and the axis of rotation.
Wι (7.5 ft) cos 53° + Wb (6 ft) cos 53° = F (15 ft) sin 53 °
Substitute the values for the weights of the ladder and the boy, respectively.
(20 lb) (7.5 ft)cos 53° + (75 lb) (6 ft) cos 53° = F(15 ft) sin 53°
Solving for F;
F = ((30 ×7.5 × cos 53°) + (75 × 6 × cos 53°))/ (15 × sin 53°)
= 33.9 lb
<u>= 33.9 Pounds</u>
