Answer:
18.3 m , 25.4°
Explanation:
d1 = 6 m, θ1 = 40°
d2 = 8 m, θ2 = 30°
d3 = 5 m, θ3 = 0°
Write the displacements in the vector form
The total displacement is given by
magnitude of resultant displacement is given by
d = 18.3 m
Let θ be the angle of resultant displacement with + x axis
θ = 25.4°
Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
.. Eq2
- Add the two equation, Eq 1 and Eq 2:
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows: