Ch 4.3 Help
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Ch 4.3 Help
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:
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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 Posts : 22
Join date : 20080917
Guestions 14
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
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
How_do_y Guest
Questions 14
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
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
How_do_y Guest
Questions 14
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
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
How_do_y Guest
Questions 14
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
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
How_do_y Guest
Question #3 correction
please note that question 3 should include a () minus sign with the answer.
elephant Guest
Question 3?
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
BGGirl Posts : 6
Join date : 20080921
question 3
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?
fresh fi Guest
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