A 95 N force exerted at the end of a 0.35 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

The angle between the wrench handle and the direction of the applied force is 26.8°

Explanation:

Given;

applied force, F = 95 N

length of the wrench, r = 0.35 m

torque on the wrench due to the applied force, τ = 15 N.m

Torque is calculated as;

τ = rFsinθ

where;

r is the length of the wrench

F is the applied force

θ is the angle between the applied force and the wrench handle

Make Sin θ the subject of the formula;

Sinθ = τ / rF

Sinθ = 15 / (0.35 x 95)

Sinθ = 0.4511

θ = Sin⁻¹(0.4511)

θ = 26.8°

Therefore, the angle between the wrench handle and the direction of the applied force is 26.8°

the answer is a 3.8cm

explanation:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.