Ch 4.3 Help

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Ch 4.3 Help

Post  Admin on Thu Sep 18, 2008 1:52 am

Question 1: (1 point)

A 4.285 kg block slides with constant velocity on a horizontal plane. The coefficient of friction is 0.2134 . The block is pushed by a force which is parallel to the plane. How large is this force (see sheet 6,30 and 24,29,29')? Indicate with a negative (positive) sign whether the direction of this force can (cannot) be obtained from the direction in which the block is sliding.

How to solve:

Question 2: (1 point)

A block slides down an inclined plane with constant velocity. The angle of inclination is 11.61 degrees with respect to the horizontal. What is the coefficient of friction (see sheet 6,30 and 27,28)? Indicate with a positive (negative) sign whether you can (cannot) calculate the normal force from the information given.

How to solve:

Question 3: (1 point)

A 37.45 N weight is suspended from two strings which both make a 17.6 degree angle with the vertical (see sheet 31; watch out! Place the angle correctly). What is the tension in each string (see sheet 32,33 but simpler). Indicate with a negative (positive) sign whether the horizontal components of the tensions in the strings cancel (do not cancel) each other.

How to solve:

Question 4: (1 point)

An object at rest at the center of a square (all sides with equal length) has 4 forces acting on it, each directed along lines from a different corner of the square toward the center. The two forces acting from the lower left and lower right corners are directed toward the center and have 96.39 N each. What is the magnitude of the two remaining forces which leaves the object at rest (see sheet 36,36')? Indicate with a positive (negative) sign whether your answer is (not) the same if the object moves with constant velocity.

How to solve:

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Guestions 1-4

Post  How_do_y on Sat Sep 20, 2008 1:41 pm

Question 1: (1 point)

A 4.285 kg block slides with constant velocity on a horizontal plane. The coefficient of friction is 0.2134 . The block is pushed by a force which is parallel to the plane. How large is this force (see sheet 6,30 and 24,29,29')? Indicate with a negative (positive) sign whether the direction of this force can (cannot) be obtained from the direction in which the block is sliding.

How to solve:
There are 4 forces acting on the object. Force (F), Force of Friction (Fr), Weight (W), and Normal (N). Net Force = 0N.
F = Fr in this case since it is moving at constant velocity.
Fr = µ*N (The coefficient of friction * the normal force)
The Normal force is the force of a surface acting against an object (for example, the table acting against the book placed on it.) Since the object is moving horizontally, N = W. And the formula for weight (W), is W=m*g.

N = W = m*g
N = W = 4.285kg * 9.81 m/s^2
N = 42.0359 N
F = Fr = µ*N
F = Fr = 0.2134 * 42.0359 N
F = 8.9705 N

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Questions 1-4

Post  How_do_y on Sat Sep 20, 2008 2:01 pm

Note for QUESTION 1, the answer is NEGATIVE.

Question 2: (1 point)

A block slides down an inclined plane with constant velocity. The angle of inclination is 11.61 degrees with respect to the horizontal. What is the coefficient of friction (see sheet 6,30 and 27,2? Indicate with a positive (negative) sign whether you can (cannot) calculate the normal force from the information given.

How to solve: (answer is negative)

Since the mass is on an incline, the N is not equal to W anymore, since N is always perpendicular to the surface.

F = Fr
F = Fr = m*g*sinθ
µ = Fr/N
N = m*g*cosθ
µ = (m*g*sinθ)/(m*g*cosθ)
µ = sinθ/cosθ
µ = sin(11.61)/cos(11.61)
µ = .2055

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Questions 1-4

Post  How_do_y on Sat Sep 20, 2008 2:10 pm

Question 3: (1 point)

A 37.45 N weight is suspended from two strings which both make a 17.6 degree angle with the vertical (see sheet 31; watch out! Place the angle correctly). What is the tension in each string (see sheet 32,33 but simpler). Indicate with a negative (positive) sign whether the horizontal components of the tensions in the strings cancel (do not cancel) each other.

How to solve:

It might be a good idea to draw a diagram for this. There is one force down (W) and two forces up (T(y)).
2*T(y) = W = 37.45
T(y) = 37.45/2
T(y) = 18.725

Now use trig to solve for T:
T = T(Y)/cos θ
T = 18.725/cos(17.6)
T = 19.6445 N

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Questions 1-4

Post  How_do_y on Sat Sep 20, 2008 2:13 pm

Question 4: (1 point)

An object at rest at the center of a square (all sides with equal length) has 4 forces acting on it, each directed along lines from a different corner of the square toward the center. The two forces acting from the lower left and lower right corners are directed toward the center and have 96.39 N each. What is the magnitude of the two remaining forces which leaves the object at rest (see sheet 36,36')? Indicate with a positive (negative) sign whether your answer is (not) the same if the object moves with constant velocity.

How to solve: (Answer is positive)

force(ur) = force(ul) = force(lr) = (force(ll)
Answer = 96.39

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Question #3 correction

Post  elephant on Sat Sep 20, 2008 11:11 pm

please note that question 3 should include a (-) minus sign with the answer.

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Re: Ch 4.3 Help

Post  BGGirl on Tue Sep 23, 2008 10:07 pm

For Quiz question 2 the Answer is negative.

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Question 3?

Post  BGGirl on Tue Sep 23, 2008 10:10 pm

Why is it that their is Two forces going [u]upward[/u] for question 3? .. I guess I am having a hard time visualizing it

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question 3

Post  fresh fi on Wed Sep 24, 2008 12:25 am

I am having a hard time seeing where the angle is going. If it is with respect to the vertical does it mean in between the two strings?

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Re: Ch 4.3 Help

Post  green on Wed Sep 24, 2008 4:33 am

For Quiz question 1 the Answer is negative.

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Re: Ch 4.3 Help

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