Given vectors and as follows, = 2i + 3j - 5k = -i + 6j + 4k Find the angle betwe

Question

Given vectors and as follows, = 2i + 3j - 5k = -i + 6j + 4k Find the angle between and 10 points If the vector in problem 1 passes through the point P(1,1,1), find the normal distance from the origin to the line of action of . 10 points Point A is fixed on the ground. Find the resultant moment about the point A. Find also the forces applied by the ground on the point A for equilibrium. 10 points If the 30N force has positive x component and |OP| = 6m, find the resultant of all the forces about the origin. 10 points The rectangular block can rotate freely about the point C. The circular ball can roll freely over the rectangular block. The weights of the block and ball are 1000N and 100N, respectively. Find the minimum force (magnitude and angle with the vertical) that has to be applied at point A to prevent the rotation of the block. Repeat the problem when the ball is at the point A. 20 points

Explanation / Answer

1.

Angle between vectors A and B = acos [(A.B) / (|A|*|B|)]

|A| = sqrt (2^2 + 3^2 + 5^2) = 6.164

|B| = sqrt (1^2 + 6^2 + 4^2) = 7.28

A.B = 2*(-1) + 3*6 + (-5)*4 = -4

theta = acos (-4 / (6.164*7.28))

= acos (-0.08913)

= 95.11 deg

Unit vector along A is n = (2i + 3j - 5k) / 6.164

Point a = (1, 1, 1) or i + j + k.

Equation of line passing through (1, 1, 1) will be

x = (i + j + k) + t*(2i + 3j - 5k) / 6.164............where t is a scalar.

Origin Point p = (0, 0, 0)

a - p = (1-0) i + (1-0) j + (1-0)k = i + j + k

(a-p).n = (i + j + k).(2i + 3j - 5k) / 6.164 = (1*2 + 1*3 + 1*(-5)) / 6.164

(a-p).n = 0

Distance from (0, 0, 0) to the line is given by Magnitude (a-p) - ((a-p).n)n

distance = Magnitude (a - p)

= sqrt (1^2 + 1^2 + 1^2)

= 1.732

2.

Net moment = 10*15 + 20*0 - 15*10 - 30*15 + 25*0 + 20*10 = -150 Nm (clockwise) or 150 Nm (anticlockwise)

Net force in x-direction = 10 - 30 +25 = 5 N

Reaction force = 5 N towards left.

Net force in y-direction = -20 + 15 - 20 = -25 N

Reaction force = 25 N upwards.

3.

x-component of 30 N force = 30*sin80 = 29.54 N

y-component of 30 N force = 30*cos80*cos50 = 3.35 N

z-component of 30 N force = -30*cos80*sin50 = -3.99 N

Net force in x-direction = 29.54 N

Net force in y-direction = 3.35 + 10 = 13.35 N

Net force in z-direction = 20 - 3.99 = 16.01 N

Resultant moment about x-axis = 0 - 13.35*(6cos38.66) + 16.01*(6sin38.66) = -2.54 Nm

Resultant moment about y-axis = 29.54*(6cos38.66) + 0 + 0 = 138.4 Nm

Resultant moment about z-axis = -29.54*(6sin38.66) + 0 + 0 = -110.72 Nm

4.

Balancing moments, 1000*20 + 100*30 + Fx*20 = 0

Fx = -1150 N

Balancing vertical direction forces, Fy - 1000 - 100 = 0

Fy = 1100 N

Net force = sqrt (1150^2 + 1100^2) = 1591.38 N

Angle with vertical = atan (1150 / 1100) = 46.27 deg

When ball is at A, balancing moments yields, 1000*20 + Fx*20 = 0

Fx = -1000 N

Fy = 1100 N

angle = atan (1000 / 1100) = 42.27 deg

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.