Question

A force of 314 N is exerted on the end of a wrench in order to apply a torque of 40.0 N·m to a bolt head. The point of application of the force is 21.5 cm from the center of the bolt. What angle does the force make with respect to the wrench handle?

Answer 1

Answer:

36.16 degree

Explanation:

We are given that

Force,F=314 N

Torque=T=40 Nm

Distance,r=21.5 cm=$21.5\times 10^{-2} m$

1 m=100 cm

We have to find the angle made by force with respect to the wrench handle.

We know that

Torque,T=$rfsin\theta$

Using the formula

$40=21.5\times 10^{-2}\times 314sin\theta$

$sin\theta=\frac{40}{314\times 21.5\times 10^{-2}}$

$sin\theta=0.59$

$\theta=sin^{-1}(0.59)$

$\theta=36.16 degree$

Answer 2

Answer:

Explanation:

Force, F = 314 N

Torque, τ = 40 Nm

distance, r = 21.5 cm = 0.215 m

Let the angle is θ.

τ = r x F x Sin θ

40 = 0.215 x 314 x Sin θ

Sin θ = 0.5925

θ = 36.3°