Figure 5-39 shows an overhead view of a 0.0250kglemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force has a magnitude of 6.00 Nand is at . Force F1has a magnitude of 7.00 N and is at θ1=30.00 . In unit-vector notation, (a) what is the third force if the lemon half is stationary, (b) what is the third force if the lemon half has the constant velocity v =(13.0i^-14.0i)m/sand(c) what is the third force if the lemon half has the varying velocity v¯=(13.0i-14.0i)m/s where tis time?

Short Answer

Expert verified
  1. F3=1.70Ni+3.06Nj
  2. F3=1.70Ni+3.60Nj
  3. F3=2.02Ni+2.71Nj

Step by step solution

01

Given information

  • First force on the body,F1=6.00N
  • Angle made by the first force, θ1=30°
  • Second force on the body, F2=7.00N
  • Angle made by the second force, θ2=30°
  • The velocity of the object is v¯=13.0i-14.0j
02

Understanding the concept

Free body diagrams represent the forces acting on the object. It can be used to find the direction and magnitude of the forces acting on the object. The net force on the object can also be found by using Newton’s law of motion which states net force is equal to the vector sum of all the forces.

Draw the free-body diagram of the block, which gives x and y components of the F1andF2. Using Newton’s second law of motion, find outF3acting on the lemon half when it is stationary. When it is moving with constant velocity, it has zero acceleration. HenceF3remains the same as the previous one. Due to varying velocities, the lemon half has acceleration. We can find acceleration by differentiatingv with respect to the time.

Formulae:

Fnet=maa=dvdt

Here, Fnetis the net force, m is the mass, a is acceleration, and v¯ is the velocity of the body.

03

Draw the free body diagram

04

a) Find the unit vector notation of the third force, when the lemon half is stationary

According to the free body diagram, the x and y components ofF1areFtx=-F1cosθ1andF1y=F1sinθ1.

The x and y components ofF2areF2x=F2sinθ2andF2y=-F2cosθ2

localid="1657091009508" F1=-F1cosθ1+F1sinθ1F2=-F2sinθ2+F2cosθ2

The lemon half is stationary hence it has no acceleration. According to Newton’s second law, the net force acting on the lemon half is zero.

Fnet=ma¯F1+F2+F3=0-F1cosθ1i+F1sinθ1j+F2sinθ2i-F2cosθ2j+F3=0

Therefore,

F3=F1cosθ1i-F1sinθ1j-F2sinθ2i+F2cosθ2j

Now, substitute the values of forces and angle in the above equation.

F3=6cos30i-6sin30j-7sin30i+7cos30j=1.70Ni+3.06Nj

Therefore, the third force acting on the object is 1.70Ni+3.06Nj

05

b) Unit vector notation of third force, when the lemon half has constant velocity v =(13.0i⏜-14.0j⏜)m/s 

When lemon half has constant velocity, so it has zero acceleration. The third force for it remains the same as,

F3=1.70Ni+3.06Nj

06

c) Unit vector notation of third force when the lemon half has the varying velocity v =(13.0i⏜ -14.0j⏜)m/s

The acceleration due to varying force is.

a=dvdt

Substitute the equation for velocity to find the acceleration.

a=d13.0ti-14.0tjdt=13.0i+14.0j

Now, use the value of acceleration in the equation of Newton’s second law to calculate the force.

Fnet=ma-F1cosθ1i+F1sinθ1j+F2sinθ2i-F2cosθ2j+F3=m13.0i+14.0j

Solve it for third force as,

F3=m13.0i+14.0j+F1cosθ1i-F1sinθ1j-F2sinθ2i+F2cosθ2j

Substitute the values of forces to calculate the third force.

F3=0.0250kg13.0i+14.0jm/s2+6cos30i-6sin30j-7sin30i+7cos30j=2.02Ni+2.71Nj

Therefore, the third force is 2.02Ni+2.71Nj

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