Question

The wheel has weight W. The minimum value of F required to pull the wheel up the step is –

• A. W cosq
• B. W tan q
• C. W sin q
• D. W cot q

Answer 1

Answer: (A)

W cosq

Answer 2

F cos a – N cos q = 0

̃ F cos a = N cos q

N sinq + F sin a – W = 0

̃ N sinq = F sina – W

F cos a tan q = F sin a – W

F (sin a – cos a tan q) = W

F = $\frac { \mathrm { W } } { \sin \alpha - \cos \alpha \tan \theta }$

F = $\frac { \mathrm { W } \cos \theta } { \sin ( \alpha - \theta ) }$

F is minimum when

a – q = 90⁰

a = 90⁰ + q

F_min = W cosq

(ii) Second method.

The wheel is in equilibrium under the action of three forces.

i.e., $\overrightarrow { \mathrm { F } }$ + $\overrightarrow { \mathrm { N } }$ + $\overrightarrow { \mathrm { W } }$ = 0

vector triangle should be drawn so that ‘F’ is minimum. This is done by drawing a perpendicular from tip of weight W on line of action of N.

From figure a + q = p/2

And F­_min = W cos q

Third method : for (ii)

Torque method

Torque is t = Fd

Where ‘d’ is perpendicular distance of line of action of force F from axis. For given torque force ‘F’ is minimum when ‘d’ is maximum.

Choose axis to be passing through ‘A’ and perpendicular to plane of figure

from above figure ‘d’ is maximum when

a + q = p/2 i.e., when AO is equal to d

i.e. d = R

Taking net torque about ‘A’ & putting it equal to zero

FR – W (R cos q) = 0

F_min = W cos q