- #1

- 12

- 0

**1. Homework Statement**

Hey I was wondering if someone would just check over this bonus question for me.

1) A father pulls his child on a sled by pulling on rope over his shoulder. He exerts a force of 50 N at an angle of 30*. a) Using the coefficient of friction for steel on ice, calculate the acceleration of the sled if its mass plus the child's is 20 kg, assuming the sled is already in motion. b) What minimum force must the father exert to start the sled moving from rest?

2) Block B in the figure (in attached file) weighs 711 N. The Coefficient of static friction between the block and the horizontal surface is 0.25. Find the maximum weight of block A for which the system will be stationary. Assume the rope attached to the wall make a 30* angle with the horizontal.

[tex]\mu[/tex]

F=ma

1) a) F

Fappcos[tex]\theta[/tex]-f = ma Fn+Fappsin[tex]\theta[/tex]= Fg

Fappcos[tex]\theta[/tex]- [tex]\mu[/tex]

a = Fappcos[tex]\theta[/tex]- [tex]\mu[/tex]

a = (50cos30 - (0.02(20(9.8)- 50 sin30))/20

a = 1.99m/s/s

b) Fcos[tex]\theta[/tex]- [tex]\mu[/tex]

Fcos[tex]\theta[/tex]- [tex]\mu[/tex]

Fcos[tex]\theta[/tex]+[tex]\mu[/tex]

F(cos[tex]\theta[/tex]+[tex]\mu[/tex]

F =[tex]\mu[/tex]

F = (0.4(20*9.8)) / (cos30 + (0.4(sin30))

F = 73.5 N

2) I was confused on this one and treated it as a hanging mass problem but here is my attempt:

Fb=ma

Fb=mg-[tex]\mu[/tex]kmg

Fb= 711 - (0.25(711))

Fb=533.25N

Tbsin[tex]\theta[/tex]= Tacos[tex]\theta[/tex]

Ta = Tbsin[tex]\theta[/tex] / cos[tex]\theta[/tex]

Ta = 533.25sin30 / cos30

Ta = 307.9N

The maximum weight for block A is 307.9N because if it were greater the string would break.

Hey I was wondering if someone would just check over this bonus question for me.

1) A father pulls his child on a sled by pulling on rope over his shoulder. He exerts a force of 50 N at an angle of 30*. a) Using the coefficient of friction for steel on ice, calculate the acceleration of the sled if its mass plus the child's is 20 kg, assuming the sled is already in motion. b) What minimum force must the father exert to start the sled moving from rest?

2) Block B in the figure (in attached file) weighs 711 N. The Coefficient of static friction between the block and the horizontal surface is 0.25. Find the maximum weight of block A for which the system will be stationary. Assume the rope attached to the wall make a 30* angle with the horizontal.

## Homework Equations

[tex]\mu[/tex]

_{s}= 0.40 [tex]\mu[/tex]_{k}= 0.02F=ma

## The Attempt at a Solution

1) a) F

_{x}= ma_{x}F_{y}=ma_{y}Fappcos[tex]\theta[/tex]-f = ma Fn+Fappsin[tex]\theta[/tex]= Fg

Fappcos[tex]\theta[/tex]- [tex]\mu[/tex]

_{k}Fn=ma Fn = Fg - Fappsin[tex]\theta[/tex]a = Fappcos[tex]\theta[/tex]- [tex]\mu[/tex]

_{k}Fn/ma = (50cos30 - (0.02(20(9.8)- 50 sin30))/20

a = 1.99m/s/s

b) Fcos[tex]\theta[/tex]- [tex]\mu[/tex]

_{s}(Fg-Fsin[tex]\theta[/tex]) = 0Fcos[tex]\theta[/tex]- [tex]\mu[/tex]

_{s}Fg +[tex]\mu[/tex]_{s}Fsin[tex]\theta[/tex])=0Fcos[tex]\theta[/tex]+[tex]\mu[/tex]

_{s}Fsin[tex]\theta[/tex])=[tex]\mu[/tex]_{s}FgF(cos[tex]\theta[/tex]+[tex]\mu[/tex]

_{s}sin[tex]\theta[/tex]) =[tex]\mu[/tex]_{s}FgF =[tex]\mu[/tex]

_{s}Fg / (cos[tex]\theta[/tex]+[tex]\mu[/tex]_{s}sin[tex]\theta[/tex]F = (0.4(20*9.8)) / (cos30 + (0.4(sin30))

F = 73.5 N

2) I was confused on this one and treated it as a hanging mass problem but here is my attempt:

Fb=ma

Fb=mg-[tex]\mu[/tex]kmg

Fb= 711 - (0.25(711))

Fb=533.25N

Tbsin[tex]\theta[/tex]= Tacos[tex]\theta[/tex]

Ta = Tbsin[tex]\theta[/tex] / cos[tex]\theta[/tex]

Ta = 533.25sin30 / cos30

Ta = 307.9N

The maximum weight for block A is 307.9N because if it were greater the string would break.