The figure shows the arrangement of given data.
The displacement is considered along X direction and time along Y direction.
The slope of the graph is given by tan θ
So tan θ = Δx/Δt = Velocity between those two points considered for finding slope of graph.
So slope of position time is graph is equivalent to kinematic quantity Velocity.
Answer:
F_ab = 260.17 N
Explanation:
Given:
- Moment of force F about CD, (M)_cd = 57 Nm
Find:
- First we will write down the position vectors of points A, B , C , D:
- We will take left and bottom most corner of cube to be the origin.
- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.
- The position vectors wrt to the origin are:
Point A = 0.2 k
Point B = 0.4 i + 0.2 j
Point C = 0.2 j + 0.4 k
Point D = 0.4 i + 0.4 k
- Now we will determine the Force vector F_ab along vector AB.
vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k
unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)
= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
Hence,
vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
- Now, form a unit vector along the line CD:
vec(CD) = D - C = 0.4 i - 0.2 j
unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)
= [sqrt(5)]*(0.4 i - 0.2 j)
- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:
vec(AC) =r_ac = C - A = 0.2 j + 0.2 k
- Now the moment about a line CD due to force is:
(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )
The cross product of r_ac and vec(F_ab) is as follows:
(M)_c = ( r_ac x vec(F_ab) ) :
(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]
The dot product of (M)_c and unit (CD) is as follows:
(M)_cd = unit(CD) . (M)_c :
(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)
(M)_cd = F_ab*(sqrt(30) / 25)
- The given magnitude of the moment is (M)_cd. Calculate F_ab:
57 = F_ab*(sqrt(30) / 25)
F_ab = 260.17 N