# Calculating Forces Homework: Net Force & Friction

• lola0706
In summary, two horizontal wires with a separation of 50 degrees are pulling on a 2180 kg mass causing it to accelerate at 0.25 m/s2 to the west. The tension in each rope is 4200 N. To find the net force on the mass, a free body diagram is used to calculate the components of each rope's force, resulting in a net force of 545 N. The calculation of friction is unclear as it is not specified which x component to use and whether to add or subtract 545. A diagram of the situation would be helpful in clarifying the solution.
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?

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

Last edited:
lola0706 said:

## 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?

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?

berkeman said:
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! :)

#### Attachments

• image-1.jpeg
27.3 KB · Views: 375
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

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

## 1. What is net force?

Net force is the overall force acting on an object, taking into account all of the individual forces acting on it. It is calculated by adding together all of the forces acting in the same direction and subtracting any forces acting in the opposite direction.

## 2. How do you calculate net force?

To calculate net force, you need to first identify all of the individual forces acting on an object. Then, add together all of the forces acting in the same direction and subtract any forces acting in the opposite direction. The resulting value is the net force.

## 3. What is friction?

Friction is the force that resists the motion of an object when it is in contact with another surface. It is caused by the roughness of the surface and the interlocking of microscopic bumps and grooves.

## 4. How do you calculate friction?

The amount of friction can be calculated using the equation F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force (the force perpendicular to the surface that the object is in contact with).

## 5. What is the coefficient of friction?

The coefficient of friction is a value that represents the amount of friction between two surfaces. It is a dimensionless quantity and is specific to the two surfaces in contact. A higher coefficient of friction means greater resistance to motion, while a lower coefficient of friction means less resistance.

• Introductory Physics Homework Help
Replies
17
Views
841
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
15
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
845
• Introductory Physics Homework Help
Replies
23
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
97
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
1K