Question

. A torque of 100Nm is required to open a bolt. The amount of force to be applied on the spanner at a length of 0.25m from the bolt is *
a) 400N
b) 4000N
c) 325N
d) 25N​

Answer 1

400 N is the required answer.

Option a is right.

GIVEN:

• Torque = 100 Nm

• Length, d = 0.25 m

TO FIND:

Force, f

FORMULA:

Torque = f × d

SOLUTION :

$torque = f \times d \\ \\ f = \frac{torque}{d} \\ \\ f = \frac{100}{0.25} \\ \\ = \frac{100}{25 \times {10}^{ - 2} } \\ \\ = \frac{100 \times {10}^{2} }{25} \\ \\ = 4 \times {10}^{2} \\ \\ f= 400 \: newton$

ANSWER:

The amount of force to be applied on the spanner is 400 N.

TORQUE:

• Torque is equal to the product of force and perpendicular distance .

• Torque measures the force required to rotate the object.

• Torque is a vector quantity.

• The vector product of distance and force gives torque vector.

• The direction of torque is perpendicular to the direction of force and distance.

• Its SI unit is Nm.

• It's dimension is [M L^2 T^−2]

Answer 2

Answer:

400 N

Explanation:

a it's spanner firce should be applied