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Zarrin [17]
3 years ago
11

A sled whose total mass with cargo is 30.0 kg rests on ice. The coefficient of static friction is 0.20 and the coefficient of ki

netic friction is 0.15 the sled is attached to a rope horizontally in which the tension force is slowly increased how much tension force applied by the rope will cause the sled to start moving?
A. 0 N
B. 9.8 N
C. 59 N
D. 290 N
Physics
1 answer:
romanna [79]3 years ago
5 0

when sled just start to move the force on the sled will be equal to the static friction

So here we need to find the value of static friction

We know that

m = 30 kg

\mu_s = 0.20

now we know that normal force on the block is counterbalanced by weight of the block

F_n = mg

F_n = 30*9.8

now in order to find the friction force we can use

F_s = \mu_s * F_n

F_s = 0.20 * 30 * 9.8

F_s = 59 N

so it requires 59 N of force to move the sled

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Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)
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Answer:

F_2=2.43\times10^{-19}\ N

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

  • m_A=200\ kg
  • m_B=250\ kg
  • m_C=1700\ kg
  • m_D=100\ kg

Now their respective coordinate positions (in cm) are as given below:

  • P_A=(0,50)
  • P_B=(0,0)
  • P_C=(-80,0)
  • P_D=(40,0)

<u>Now force on B due to A:</u>

F_{BA}=G\times \frac{200\times 250}{50^2}

F_{BA}=20G\ N

<u>Now force on B due to C:</u>

F_{BC}=G\times \frac{1700\times 250}{80^2}

F_{BC}=66.406G\ N

<u>Now force on B due to D:</u>

F_{BD}=G\times \frac{100\times 250}{40^2}

F_{BD}=15.625G\ N

<em>We observe that the forces due to masses  C&D act opposite in direction.</em>

<u>So, the net force in the x-direction:</u>

F_x=F_{BC}-F_{BD}

F_x=66.406G-15.625G

F_x=50.781G\ N in the positive x-direction

<em>We have only one force in y-direction due to mass A.</em>

So,

F_y=20G\ N in the positive y-direction.

<u>Now the net force:</u>

F_2=\sqrt{F_y^2+F_x^2}

F_2=\sqrt{(20G)^2+(50.781G)^2}

F_2=54.5776G^2\ N

F_2=2.43\times10^{-19}\ N

<u>Now the direction of this force with respect to x-axis:</u>

tan\ \theta=\frac{F_y}{F_x}

tan\ \theta=\frac{20G}{50.781G}

\theta=21.496^{\circ} in the anticlockwise direction from the positive x- axis.

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