Question

A wheel of radius 3 m is mounted on its axis. A boy pushes with a force of 50 N as shown. If cos20° = 0.9, then the torque applied by boy about its axis is​

Answer 1

Step-by-step explanation:

Torque is defined as the cross product of Vector R and Force F, where vector R represents the distance from Axis of rotation to point of application of force.

In terms of Formula, we can write it as:

→ τ = r × F

→ τ = r.F.SinФ

where, Ф is the angle between F and r vectors.

According to the question, we are given that,

→ Magnitude of Force (F) = 50 N

→ Distance from Axis (r) = 3 m

Rearranging the vectors, we get the angle between them to be 70°

Therefore substituting in the formula. we get:

→ τ = 50 × 3 × Sin 70°

According to trigonometric identity,

→ SinФ = Cos( 90 - Ф)

→ Sin 70 = Cos ( 90 - 70 ) = Cos 20

Therefore Sin 70 = Cos 20 = 0.9  [Given]

Hence Torque is calculated as;

→ τ = 50 × 3 × 0.9

→ τ = 135 N.m

Hence the torque applied by the boy along the axis is 135 N.m

Answer 2

Answer:

maybe one of this

Step-by-step explanation:

135 Nm

140 Nm

40 Nm

Zero