Answer:
(a) reaction at each front wheel is 5272N (upward)
(b) force between boulder and pallet is 4124N (compression)
Explanation:
Acceleration of the truck = 1 m/ (to the left)
when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,
= 0.5 m/ (upward) , = 0.5 m/ (upward)
Let T be tension in the cable
pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( + )g = ( + )
= 2T- (400 + 50)*(9.81 m/) = (400 + 50)*(0.5 m/)
T = 2320N
Truck: = ∑()eff: = (3.4m) + (2.0m) - T (0.6m)= (1.0m)
Nf = (2.0m)(2000 kg)(9.81 m/ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/) = 11541.2N - 409.4N - 588.2N = 10544N
∑fy (upward) = ∑(fy)eff: + - g = 0
10544 + - (2000kg)(9.81 m/ ) = 0
= 9076N
∑fx (to the left) = ∑(fx)eff: - T =
= 2320N + (2000kg)(9.81 m/ ) = 4320N
(a) reaction at each front wheel:
1/2 (upward): 1/2 (10544N) = 5272N (upward)
(b) force between boulder and pallet:
∑fy (upward) = ∑(fy)eff: + g -
= (400kg)(9.81 m/) + (400kg)(0.5 m/) = 4124N (compression)
Answer: The coefficient of expansion for the wire .
Original length of the wire = L= 410.0000 m
Initial temperature =
Final Temperature =
Increase in Length of the wire =
The coefficient of expansion for the wire .
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C