Calculating Forces Homework: Net Force & Friction

[lola0706]

Homework Statement

two horizontal wire which are separated by 50 degrees are pulling on a 2180 kg mass and cause it to accelerate to the west at 0.25 m/s2. the tension in each rope is 4200.

a) find the net force on the mass
b) how much friction is acting on the mass?

Homework Equations

fnet= ma
cos law
sine law

f = fah-fnet

The Attempt at a Solution

so i drew out a free body diagram and calculated the components.
x1 = cos25 (4200) = 3806 N
y1 = sin25 (4200) = 1775 N

x2 = the same as x1
y2 = the same as y1

i found fnet by going mass * acceleration and got 545 N

now calculating friction is difficult. i do not know if i am supposed to use only one x component or both. am i supposed to add 545 or subtract it? all help is appreciated. thanks

 

[berkeman]

Homework Statement

two horizontal wire which are separated by 50 degrees are pulling on a 2180 kg mass and cause it to accelerate to the west at 0.25 m/s2. the tension in each rope is 4200.

a) find the net force on the mass
b) how much friction is acting on the mass?

Homework Equations

fnet= ma
cos law
sine law

f = fah-fnet

The Attempt at a Solution

so i drew out a free body diagram and calculated the components.
x1 = cos25 (4200) = 3806 N
y1 = sin25 (4200) = 1775 N

x2 = the same as x1
y2 = the same as x2

i found fnet by going mass * acceleration and got 545 N

now calculating friction is difficult. i do not know if i am supposed to use only one x component or both. am i supposed to add 545 or subtract it? all help is appreciated. thanks

Welcome to the PF.

Is there a figure/diagram that goes with this question? Can you scan it or copy/paste it as an attachment?

 

[lola0706]

Welcome to the PF.

Is there a figure/diagram that goes with this question? Can you scan it or copy/paste it as an attachment?

this is what i assume the diagram would look like

the 25 degrees would be the 50 degrees shared by the 2 triangle components

oh and thanks for the welcome! :)

 

[Chestermiller]

Let x be the direction in which the mass is moving. What is the component of force exerted by each of the ropes in the x direction? What is the sum of these two components of force? Write a force balance for the mass in the x direction that includes the two components of rope force, the frictional force, and the mass times acceleration.

Chet

 

[HalfLostAndSearching]

I would like to clarify a few things before providing a response. Firstly, it is important to note that the homework statement does not specify the units for mass and acceleration, so I will assume that they are in kilograms (kg) and meters per second squared (m/s2), respectively. Secondly, the statement does not mention the direction of the acceleration, so I will assume it is in the westward direction as given in the problem.

Now, to answer the first part of the homework, we can use the equation Fnet = ma to calculate the net force on the mass. From the given information, we know that the mass is 2180 kg and the acceleration is 0.25 m/s2 to the west. Therefore, the net force on the mass is:

Fnet = (2180 kg)(0.25 m/s2) = 545 N

This means that the two horizontal wires combined are exerting a force of 545 N on the mass in the westward direction.

For the second part of the homework, we need to calculate the friction force acting on the mass. To do this, we first need to determine the forces acting on the mass in the horizontal direction. From the free body diagram, we can see that there are two forces acting on the mass in the horizontal direction: the tension force in the ropes and the friction force. We can use the cosine law to find the magnitude of the tension force in each rope (which is the same for both ropes):

T = √(x12 + y12) = √[(cos25)(4200 N)2 + (sin25)(4200 N)2] = 4470 N

Now, to find the friction force, we can use the equation Fnet = ma again, but this time in the horizontal direction. The net force in the horizontal direction is the difference between the tension force and the friction force, so we have:

Fnet = T - f = ma

Rearranging this equation, we can solve for the friction force:

f = T - ma = (4470 N) - (2180 kg)(0.25 m/s2) = 4010 N

Therefore, the friction force acting on the mass is 4010 N in the direction opposite to the acceleration (eastward).

I hope this explanation helps with your understanding of the problem. It is important to carefully consider all the information given