Question

A string is pulled with a force F = 130 N as shown. The force vector comes out to be F=10(ai+Bj+ck) N. Write B+a+c in OMR sheet.​

Answer 1

Answer:

hope it's helpful to you

Answer 2

Answer:

The value of a= 3 units, b = 12 units, c = 4 units.

Value of a + b + c = 19 units.

Explanation:

Given that :

The magnitude of vector F = 130 N

Force Vector F = 10(ai+ bj +ck).

Let the component of force vector F

• along x-axis be a i ,
• along y-axis be b j ,
• along z-axis be c k.

From geometry of figure it is clearly evident that:

• a along x-axis is (5-2) m = 3 m.
• b along y-axis is 12 m.
• c along z-axis is (6-2) = 4 m.

• Sum of these three components will form the Force Vector F and resultant of these three components will give the magnitude of Force Vector F.
• Magnitude of the Force Vector F is given by

$\sqrt{ {a}^{2} + {b}^{2} + {c}^{2} }$

Putting values of a, b, c in above equation, we get

magnitude of Force Vector F

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ {3}^{2} + {12}^{2} + {4}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: = \sqrt{9 + 144 + 16} \\ \: \: \: \: \: = \sqrt{169} \\ = 13 \:$

• Again, it is given that

$130 = \sqrt{ {(10)}^{2} ({a}^{2} + {b}^{2} + {c}^{2}) } \\ 13 = \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} }$

• Comparing both the above equations, we can conclude that:

a = 3 m

b = 12 m

c = 4m

• Hence, value of a + b + c = 19 m.