Question

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizontal.
How much of the worker’s force is not used to move the crate?

a. 200.69 N
b. 419.95 N
c. 357.71 N
d. 191.32 N

Answer 1

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

Answer 2

Answer:

The worker’s force is 200.69 N.

(a) is correct option.

Explanation:

Given that,

Force = 245 N

Angle = 55°

According to component of force

Horizontal component of force is in direction of displacement which is used to move the crate.

Vertical component of force is not used to move the crate.

We need to calculate the worker’s force which is not used to move the crate

Using y- component of force

$F_{y}=F\sin\theta$

Put the value into the formula

$F_{y}=245\times\sin55$

$F_{y}=200.69\ N$

Hence, The worker’s force is 200.69 N.