Find the value of ##θ## and frictional force acting on the block.

[chwala]

Homework Statement

A block of wood of mass $4.5$ kg rests on a table. A force of magnitude $35$ N, acting upwards at an angle $θ^0$ to the horizontal, is applied to the block but does not move it. Given that the normal contact force between the block and the tablke has magnitude $30$ N. Calculate

a. the value of $θ$.

b. the frictional force acting on the block.

Relevant Equations

Mechanics

Refreshing on this...This is a relatively new area to me ...my solutions should be correct...what i am trying to engage on is if there are any other ways of looking at the same problem.

Ok in my working i have,

$30 + 35 \cos β-45=0$

$\cos β=0.4286$

$β = 64.62^0, ⇒θ=25.4^0$ to one decimal place.


for part b,

$35\cos 25.4^0 -F=0$

$F=31.6 N$

any insight is welcome...

 

[SammyS]

Homework Statement: A block of wood of mass $4.5$ kg rests on a table. A force of magnitude $35$ N, acting upwards at an angle $θ^0$ to the horizontal, is applied to the block but does not move it. Given that the normal contact force between the block and the tablke has magnitude $30$ N. Calculate

a. the value of $θ$.

b. the frictional force acting on the block.
Relevant Equations: Mechanics

Refreshing on this...This is a relatively new area to me ...my solutions should be correct...what i am trying to engage on is if there are any other ways of looking at the same problem.

Ok in my working i have,

$30 + 35 \cos β-45=0$

$\cos β=0.4286$

$β = 64.62^0, ⇒θ=25.4^0$ to one decimal place.


for part b,

$35\cos 25.4^0 -F=0$

$F=31.6 N$

any insight is welcome...

In which direction is the frictional force exerted on the block? To the left? Or is it directed to the right?

( I hate those little Image thumbnail images ! )

( I hate emojis too ! ) Beat y'all to it !

 

[chwala]

In which direction is the frictional force exerted on the block? To the left? Or is it directed to the right?

View attachment 340430

( I hate those little Image thumbnail images ! )

( I hate emojis too ! ) Beat y'all to it !

@SammyS i think the arrow ought to point downwards; not upwards as shown in my diagram. I need to understand the 'english' used in these kind of questions... should be to the left.

 

[haruspex]

In which direction is the frictional force exerted on the block? To the left? Or is it directed to the right?

Why does it matter?

i think the arrow ought to point downwards

No, your diagram is fine. Are you posting just for confirmation or do you have some reason to think your answer is wrong?
Are you supposed to use 10m/s² for g?

I note the question specifies $\theta°$ (which I dislike). That means $\theta$ is just a number, 25.4.

 

[chwala]

Why does it matter?

No, your diagram is fine. Are you posting just for confirmation or do you have some reason to think your answer is wrong?
Are you supposed to use 10m/s² for g?

I note the question specifies $\theta°$ (which I dislike). That means $\theta$ is just a number, 25.4.

@haruspex like I indicated...I am posting for insight if any that's all...this is a new area for me...

Like the insight given on the diagram (direction of frictional force) for instance...those are the areas that may require your indulgence...

 

[kuruman]

Like the insight given on the diagram (direction of frictional force) for instance.

To figure out the direction of the friction, look ahead. You have 4 forces

1. Pulling force $F$ which is up and to the right
2. Weight $mg$ which is straight down
3. Normal force $N$ which is straight up
4. Friction $f$ which is parallel to the surface either to the left or to the right.

Looking ahead, you want the sum of the 4 forces to be zero. Since the unknown force is horizontal, the first three forces will have no up-down component but a component to the right due to the pulling force. Therefore, the force of friction must be equal to the horizontal component of the pulling force and point to the left.

Also, just to streamline your work, you don't really need angle $\beta$. Not what you did is incorrect, but you can write more simply
$30 + 35 \sin \theta - 45=0 \implies \sin\theta=\frac{3}{7}$ and
$\cos\theta=\sqrt{1-\sin^2\theta}=\frac{2\sqrt{10}}{7}.$
Then
$f=F\cos\theta=10\sqrt{10}~$N.